1. There are 16 teams divided in 4 groups.

Every team from each group will play with each other once.
The top 2 teams will go to the next round and so on the top two teams will play the final match.
Minimum how many matches will be played in that tournament?

a)43 b)40 c)14 d)50

In each group, total matches played = 4C2 = 6.

So total matches played in the first round = 6 × 4 = 24
Now top two teams from each group progress to the next round. Now these 8 teams are pooled into 2
groups. Total matches played in the second round = 6 × 2 = 12
Now 4 teams progress to the next round. Total matches played in the third round = 6
From this round, 2 teams progress to the next round. And final will be played between them.
Total matches = 24 2+ 12 + 6 + 1 = 43

2. There are 20 persons sitting in a circle. In that there are 18 men and 2 sisters. How many
arrangements are possible in which the two sisters are always separated by a man?

a)18!*2 b)17! C)17*2! D)12

Let the first sister name is A. Now she can sit any where in the 20 places (Symmetrical). Now her sister B
can sit to her left or right in 2 ways. Now the remaining 18 persons can be sit in 18 places in 18! ways.
Total = 18! × 2

3. A bag contains 8 white and 3 blue balls. Another bag contains 7 white and 4 blue balls. What is the
probability of getting a blue ball? a.3/7 b.7/22 c.7/25 d.7/15

White Blue Total

8 3 11

7 4 11

15 7 22

7/22

4. There are 2 bags. A bag contains 5 white and 10 red balls. Another bag contains 10 white balls and 7
red balls. what is the probability of taking the red ball from one of the bag? a.55/102 b.17/21 c.15/17
d.7/8

First bag= 10 red balls out of 15 balls

½ ( 10/15)+ ½ (7/17)

½(2/3) + ½ (7/17)

1/2[2/3+7/17]
½[34+21/51]

½(55/51)

55/102

5. In a 3*3 square grid comprising 9 tiles each tile can be painted in red or blue color. When the tile is
rotated by 180 degree, there is no difference which can be spotted. How many such possibilities are
there? a.16 b.32 c.64 d.256

6. On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before
he throws 7 is? a.40% b.45% c.50% d.60%

To get 5

(2,3), (3,2), (4,1), (1,4) => 4 chances

To get 7

(1,6),(6,1),(5,2),(2,5),(4,3),(3,4)=> 6 chances

Total Chances=10

Probability = 4/10 or 40/100

40%

6. 3 dice are rolled. What is the probability that you will get the sum of the no’s as 10?

a. 27/216 b.25/216 c.10/216 d.1/11

Three dices so that total throws = 6*6*6 =216

D1 D2 D3

6 3 1 3! =6

6 2 2 3!/2!=3

5 4 1 3!=6

5 3 2 3!=6

4 4 2 3!/2!=3

4 3 3 3!/2!=3

6+3+6+6+3+3=27 then 27/216

7. How many 6 digit even numbers can be formed from digits 1 2 3 4 5 6 7 so that the digit should not
repeat and the second last digit is even? a) 6480 b)320 c)2160 d)720

5 *4 *3 *2 * 2* 3=720

8. How many vehicle registration plate numbers can be formed with digits 1,2,3,4,5(no digits being
repeated)if it is given that registration number can have 1 to 5 digits ? a.205 b.100 c.325 d.105

5*4*3*2*1=120 ; 5*4*3*2=120 ; 5*4*3=60 ; 5*4*=20 ; 120+120+60+20+5= 325

9. A Box has 13 white chips, 7 blue chips and 6 green chips. What is the probability that, if two chips are
drawn from the box in succession, one is blue and other is white. a.7/50 b.20/26 c.4/20 d.7/25
13
c1/26c1 * 7c1/25c1

13/26*7/25 =7/50

10. How many possible ways can you write 3240 as a product of 3 positive integers a, b,c?

a.320 b.450 c.480 d.360

11. in how many ways a team 11 must be selected from 5men and 11 women such that team must
comprise of not more than 3men? a.3345 b. 2256 c. 4000 d. 3456

(5c0*11c11)+(5c1*11c10)+(5c2*11c9)+(5c3*11c8)

=2256ways

12.
There are 6 red shoes & 4 green shoes. If two of red shoes are drawn what is the probability of getting red
shoes

taking 2 red shoe the probablity is 6C2

from 10 shoes probablity of taking 2 red shoe is 6C2/10C2=1/3

The probability of two events A and B are0.25 and 0.40 respectively. The probability that both A and B
occur is 0.15. The probability that neither A nor B occur is _________

we are apply that formula..............

P(AorB)=P(A)+P(B)-P(A AND B)
=.25+.40-.15
=.50
but the probability of neither A nor B=1-.50
=.50
so correct ans is .50

Three cards are drawn at random from an ordinary pack of cards. Find the probability that they will
consist of a king, a queen and an ace.

From 4 kings, draw a king]*

[from 4 queens, draw a queen]*
[from 4 aces, draw an ace] /
[from 52 cards, draw any 3]
= 4C1*4C1*4C1/ 52C3
= 16/5525 or 0.2896 %

Two dice are tossed. The probability that the total score is a prime number is:

A. 1/6 B. 5/12 C. ½ D. 7/9

In Simultaneous Throw Of Two Dice,n(S)=6*6=36

let E=Event Of Getting A Prime
Number={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}
P(E)=n(E)/n(S)
=12/36
=5/12
Two socks are to be picked at random from a drawer containing
only black and white socks. What is the probability that both are
white?
probability that both are white =1/4
as one favourable option (WW) out of four possibilities (WW,WB,BW,BB).

Q. A coin is so unbalanced that it may come both heads in 2 tosses

as it may come tails in a single toss. What is the probability of
getting a head in a single toss.

ans:1/2
in two toss both heads means=1/2 * 1/2=1/4
similarly in single toss probability of getting head is 1/2

There are 9 balls of which one is defective. What are the minimum
number of chances required to find the defective one?

Some children goto ice-cream shop. 9 flavours are available there. Each child takes a
cone with two different flavours. No two children take same combination and they have
taken all such possible combinations. How many children went to ice cream shop?

there are 9 flavours ,each one takes 2

therefore 9c2 by using propability
9*8/2*1=36
so childrens=36

If from a deck of 52 cards,4 cards are selected and 1 card of it

should be spade and another should be heart,in how many ways can
these cards be selected?
a)13^2*50C2 b)52C4 c)26*50C2 d)13C4

Option a) 13^2*50C2
13C1 spades
13C1 hearts
remaining 2 cards will be choosen in 50C2 ways..

There are 6 cities, of which each is connected to every other city. How many
different routes can one trace from A to B, such that no city is touched more
than once in any one route

c) 65
direct=1
if 1city is covered=4
if 2CITies are covered=4*3=12
if 3cities are coverd =4*3*2=24
if 4 cities are covered=4*3*2*1=24
total=65

r = number of flags;
n = number of poles;
Any number of flags can be accommodated on any single pole.
r= 5 n = 3 . If first pole has 2 flags ,third pole has 1 flag, how many
ways the remaining can be arranged?

it can be arranged in 6 ways

2 flags are remaining and as it is given that 3 poles can have any no. of flags therefore
3p2=3!/(3-2)!
=3*2=6 ways

In how many ways 8 members can be selected for a team such that women constitute
at least 50% of the team from a group of 8 men and 8 women.

in all we have to choose 8 members from total

where at least 50% women are present
so we have cases like
4 women and 4 men = 8C4 * 8C4 = 70*70=4900
5 women and 3 men = 8C5*8C3 = 56*56=3136
6 women and 2 men = 8C6*8C2=28*28=784
7 women and 1 man = 8C7*8C1=8*8=64
8 women and 0 man = 8C8 =1
adding all the 4 values above obtained we have
8885 ways

a box contains 12 marbels of 3 different colours green,yellow and

blue 4each,if you were to close your eyes and pick them at random
how many marbels must you take out to be sure that there are at
least two of each colour among the marbels you picked out?

ans: 160600 (ie 6C3 * 14C6)

check it..

Ravi has a bag full of 10 Nestle and 5 Cadbury chocolates. Out of

these, he draws two chocolates. What is the probability that he
would get at least one Nestle Chocolate ?
1) 19/21 2) 3/7 3) 2/21 4) 1/3

Total number sample space two chocolates can be drawn in the way of n(S) = 15C2= 105
At least one to be nestle.. n(E) = 10C1 X5C1 + 10C2 X 5C0 = 95
P(E) = 95/105 = 19/21

A survey of a village showed that 1 / 10 of the total population speak neither

Hindi nor English. 1 / 5 of them cannot speak English and 3 / 7 of them cannot
speak Hindi. What percentages of people know only one language?
A) 42.86 % B) 40 % C) 35 % D) None of these

persons who can speak English = 1 - (1/5) = 4/5

persons who can speak Hindi = 1 - (3/7) = 4/7
persons who can speak any or both = 1 - 1/10 = 9/10
so total persons speak both = 4/5 + 4/7 - 9/10 = 33/70
persons speak any = 1 - 33/70 - 1/10 = 3/7
% = 300/7 Answer is A

varun is guessing which of the 2 hands holds a coin. what is the probability
that varun guesses correctly three times in a row?
a. 1/6 b. 1/2 c. 1/4 d. 1/8 ½*1/2*1/2

A fair coin is toss repeatedly. If head appears on first 4 tosses then what is
probability of appearance of tail in fifth toss-
a) 1/5 b) 2/5 c) ½ d) 4/5 e) None ½

it is a small town railway station and there are 25 stations on that line.at each
of the 25 stations the passengers can get tickets for any other 24 stations.how
many different kind of tickets do you think the booking clerk has to keep?

600 different kind of tickets booking clerk has to keep.

24 type of tickets are to be issued from each originating station.

Therefore, 25*24 = 600 tickets

find out the probability to get 1Rs coin head, 2Rs coin Tail and 5Rs coin head??
a) ½ b) 1/8 c) 1/16 d) None of these

1Rs coin head=1/2

2Rs coin Tail=1/2
5Rs coin head=1/2

=(1/2)*(1/2)*(1/2)
=1/8

A quiz has one mcq question with a,b and c as options. and two questions with
true/false answers. what is the probability of giving all 3 answers correct?

1/12

probability of getting mcq answer correct = 1/3

probability of getting one true/false answer correct = 1/2

probability of getting all 3 answers correct = 1/3 * 1/2 *1/2

=1/12

Planet fourfi resides in 4-dimensional space and thus the currency

used by its residents are 3-dimensional objects. The rupee notes are
cubical in shape while their coins are spherical. However the coin
minting machinery lays out some stipulations on the size of the
coins. The diameter of the coins should be at least 64mm and not
exceed 512mm. Given a coin, the diameter of the next larger coin is
at least 50% greater.
The diameter of the coin must always be an integer.
You are asked to design a set of coins of different diameters with
these requirements and your goal is to design as many coins as
possible. How many coins can you design? 8 5 6 9

ans is 6

this is a series where t(n+1)=t(n)*1.5

64,96,144,216,324,486

If the letters of the word CHASM are rearranged to form 5 letter words such
that none of the word repeat and the results arranged in ascending order as in
a dictionary what is the rank of the word CHASM?

Solution
total words = 120
now rank of CHASM will be
120*(2-1)/5 + 24*(2-1)/4 + 6*(1-1)/3 + 2*(2-1)/2 + 1 = 24 + 6 + 0 + 1 + 1 = 32

25 buses are running between two places P and Q. In how many

ways can a person go from P to Q and return by a different bus?
1) 625 2) 600 3) 576 4) None of these 25*24

There are 6 periods in each working day of a school. In how many ways can
one organize 5 subjects such that each subject is allowed at least one period?
1) 3200 2) 3600 3) 2400 4) None of these

5 subjects can be arranged in 6 periods in 6P5 ways.

Remaining 1 period can be arranged in 5P1 ways.

Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid
overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800

8 chairs are numbered from 1 to 8. Two women first choose two

chairs from those marked 1 to 4 and 3 men selects 3 chairs from
the remaining. Find the number of possible arrangements

1,2,3,4,5,6,7,8 are 8 chairs.

two women first choose two chairs from 1,2,3,4

in 4p2 ways = 12 different ways

3 men selects 3 chairs from the rest 6 chairs = 6p3 = 6*5*4 = 120.
Total number of possible arrangement = 12*120 = 1440.

How many 6 digit even numbers can be formed from digits 1 2 3 4 5 6 7 so

that the digit should not repeat and the second last digit is even?
a)6480 b)320 c)2160 d)720

there are six digits..

a b c d e f(say)

f=possible digits are 3 -(2,4,6)

e=possible 2 digits - (2,4,6)

and then rest...

d=possible 5 digits

c=possible 4 digits

b=possible 3 digits

a=possible 2 digits

the digits goes on decreasing since digits are not supposed 2 b repeated...
hence ans=>3*2*5*4*3*2=720...:)

3 persons A, B and C are standing in a queue. There are 5 persons between A and B
and 8 persons between B and C. If there are 3 persons ahead of C and 21 persons
behind A, what could be the minimum number of persons in the queue ?
Option
1) 41 2) 40 3) 28 4) 27

arrangement is like this

B
5 person
A
3 person
c
17person
plz include A,B,C while calculating answer 1+5+1+3+1+17
so answer is 21+A+5+B which means 21+1+5+1=28

How many 6 digit telephone numbers can be formed if each number

starts with 35 and no digit appears more than once?
1) 720 2) 360 3) 1420 4) 1680

we have to make 6 digit number...from which 2 digits are already given.so we are left with

4 digits and numbers from 0-9(exept 3 & 5)

so we have to find number of ways to arrange 8 numbers (i.e. 0,1,2,4,6,7,8,9) in to make 4

digit number..so that 8P4=8! / 4! = 1680.

There are 6 tasks and 6 persons

Task 1 cannot be assigned either to person 1 or person 2
Task 2 must be assigned to either person3 or person 4.
Every person is to be assigned one task.
In how many ways can be assignment done?
a)192 b) 360 c) 144 d) 180

3C1.2C1.4C1.3C1.2C1.1C1=144 ansr
2nd task = 2C1
1st task (6-2-1(2nd task))=3C1
3rd task (6-2(1st and 2nd tasks))=4C1
4th task (6-3(1st,2nd,3rd tasks))=3C1
5th task (6-4)=2C1
6th=1C1
A 3*3 grid is coloured using red and blue colours,such that if we rotate the grid about
its centre plane by 180 degrees,the grid looks the same.The number of ways to colour
the grid this way is:a)256 b)64 c)16 d)32

2*2*2*2=32,because if opposite faces are painted by either blue or red and rotated by 180 degrees
each is rotated by 2 faces so 2*2*2*2=32(since centre face remains constant and remaining 8 are
rotated)

A girl has to make pizza with different toppings. There are 8 different toppings. In how
many way can she make pizzas with 2 different toppings?
a)16 b)56 c)112 d)28
it can be done in 8c2 ways=28
ans:28

Zada has to distribute 15 choclates among 5 of her children

sana,ada,amir,farhan. she has to make sure that sana get atleast 3
and atmost 6 choclates.in how many ways can this be done?

there would be 4 cases

1.A gets 3 chocolates then the other 12 can be distributed among other 4 children in (12-1)C(4-
1)ways ,ie;165 ways
2.A gets 4 chocolates then the other 11 can be distributed among other 4 children in (11-1)C(4-
1)ways ,ie;120 ways
3.A gets 5 chocolates then the other 10 can be distributed among other 4 children in (10-1)C(4-
1)ways ,ie;84 ways
4.A gets 6 chocolates then the other 9 can be distributed among other 4 children in (9-1)C(4-1)ways
,ie;56 ways
adding all above cases,we get 165+120+84+56=425 ways.....
hence the solution is 425 ways.

in how many ways 10 identical looking pencils to 4 students so that each student get
atleast one pencil

firstly give 1 pencil each to the 4, now we can distribute the remaining 6 pencils any way we like.
using the "stars and bars" formula, (6+4-1)C(4-1)
= 9C3 = 84
15 people sit round a circular table what are the odds against two particular people
sitting together. (a)7:1 (b)6:1 (c)5:1 (d)1:5
ans is 6:1
odds agianst an event=
(possibilities of event not happenning)/(possibilities of event happening)
=possibilities that 2 people are not togeather/possi that they are togeather
= (14!-2*13!)/(2*13!)
= 6:1

In how many different ways can 5 girls and 5 boys form a circle
such that the boys and the girls alternate?
1) 2880 2) 1400 3) 1200 4) 3212

the answer is 2880

explanation : since in circular permutation no. of ways person can be arrange is (n-1) !
now suppose the girls are seated in !4 ways now there are 5 spaces created (try to visualise ) now
these can be filled in !5 ways so the ans is !4*!5=2880

20 passengers are to travelled by a doubled decked bus which can accommodate 13 in

the upper deck and 7 in the lower deck. The number of ways that they can be
distributed if 5 refuse to sit in the upper deck and 8 refuse to sit in the lower deck is:
(a) 25 (b) 21 (c) 18 (d) 15

refused to sit on upperdeck --->5

refused to sit on lowerdeck --->8

total people remaining to arrange is = 20-13=7

remaining place in upper deck = 5

therefore, no. of ways possible= 7c5 = 21

remaining place in lower deck = 2

therefore, no. of ways possible= 7c2 = 21,also

no. of ways possible is 21.

From a collection of 12 bulbs of which 6 are defective 3 bulbs are

chosen at random for three sockets in a room. Find the probability
that the room is lighted if one bulb is sufficient to light the room.
1-(6C3/12C3) = 10/11

A man has 3 shirts, 4 pairs of trousers and 6 ties. What are the number of
ways in which he can dress himself with a combination of all the three?
A. 72 B. 13 ! / ( 2 ! * 4 ! * 6 !) C. 3 ! * 4 ! * 6 ! D. 13

3c1 * 4c1 *6c1 = 3*4*6 =72

There are two boxes. one containing 32 red balls and other containing 31green
balls. you are allowed to move the balls between the boxes so that when you
choose a box at random and a ball at random from the chosen box ,the
probability of getting a red ball is maximized. The maximum probability is
a.0.25 b.0.51 c.0.75 d.0.50

Move 31 balls from red ball box to other box having 31 green balls.

now one box has one red ball and other box is having 31 red balls and 31 green balls.
then prob of getting red ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4

A teacher set a ques ppr consist of 8 question of total 40 marks such that each
question contains minimum two marks and integer number of marks. In how
may ways this can be done? a)48C7 b)31C7 c)32C7 d)40C8

2no is sure to divide b/w 8 ques so no which we need to divide is only=40-(8×2)=24

and by the therom if we want to divide n different thing in r part then formula is=n+r-1Cr-

1
So ans is =24+8-1C8-1=31C7

In how many ways can 7 distinct objects be divided among three people so that either
one or two of them do not get any object?

Case 1: Two guys don't get anything. All items go to 1 person. Hence, 3 ways.
Case 2: 1 guy doesn't get anything. Division of 7 items among 2 guys. Lets say they are called A & B.

a) A gets 6. B gets 1. 7 ways.

b) A gets 5. B gets 2. 21 ways.
c) A gets 4. B gets 3. 35 ways.
d) A gets 3. B gets 4. 35 ways.
e) A gets 2. B gets 5. 21 ways.
f) A gets 1. B gets 6. 7 ways.

Total = 126 ways to divide 7 items between A & B.

Total number of ways for case 2 = 126*3 = 378 (Either A, or B, or C don't get anything).

Answer = Case 1 + Case 2 = 3 + 378 = 381

assume there are three persons A,B ,C

now consider following two cases
1.
A do not get anything
in this case 7 objects can be distributed between B and C as follows
BC
1 6 - total ways are 7!/1!*6!=7
2 5 -total ways are 7!/2!*5!=21
3 4 - total ways are 7!/3!*4!=35
4 3 - total ways are 7!/4!*3!=35
5 2 - total ways are 7!/5!*2!=21
6 1 - total ways are 7!/6!*1!=7
hence total ways of distributing objects between B and C = 126
similarly for (A c)and(AB) each =126
CASE2-
two persons do not get anything
in thi case there will be three waya these are
007 ,070 ,700
combining both the cases
total no of ways =126*3+3=381

A person has in is bag 14 notes of Rs. 10 each, 9 notes of Rs. 5

each, 4 notes of Rs. 2 each and 7 notes of Re 1 each. In how many
different ways can be contribute to a charitable find?
a) 599 b) 5999 c) 3528 d) 600
As per the questions,
14 notes of Rs10...so person can contribute
Rs10 note in 14 ways.if person dont give Rs
10 note then number of ways of
contribution are 14+1=15.
9 notes of Rs 5...so person can contribute
Rs5 note in 9 ways.if person dont give Rs
10 note then number of ways of
contribution are 9+1=10.
and so on..
SO the number of ways can be contribute
to a charitable fund
15*10*5*8-1=5999(option b) answer
here -1 is used because person must have
to use atleast one note.

In how many ways 3 boys can be placed in 9 seats in a single row,such that no two
boys are adjacent to each other.
they can be either arrange at odd places or at even places
1:if they are arranger at odd (1,3,5,7,9)5p3
2:if they are arranged at even places(2,4,6,8)4p3
so 5p3+4p3=54

In a pizza restaurant, you get a basic pizza with two toppings:

cheese and tomato. You can also make up your own pizza with
extra toppings. You can choose from 6 different toppings, e.g.
olives, ham, mushoroom, salami etc Ross wants to order a pizza
with 2 different extra toppings. How many different combinations
are possible?

6c2=15
bcozz
we have to select two topppings from 6 toppings and here order is not important so jst 6c2

If a traveller was offered 5 city in asia and 11 city in europe and asked to chose atleast
3 city in asia and overall a 11 city. How many ways can he travel
its right

no. of ways = 5C3*11C8 + 5C4*11C7 + 5C5*11C6

There is a row of 50 children of which 2 of them are girls.They are standing in an initial
order such that a set of 6 boys, who are armature astronomers, stand next to each
other in a certain order. Now the positions of boys and girls are shuffled such that ,the
order of girls remain the same the amature astronomers are together in the same
order, but the position and order of remaining boys are changed.How many distinct
arrangements of the students can be done (including the initial order) by preserving the
order of girls and keeping the amature astronomers together .
(a) 44 ! (b) 44 ! * 946 (c) 43 ! * 946 (d) 43 !
A mother with 7 children takes three at a time to a cinema. She goes with every group
of three that she with distinct can form. How many times can she go to the cinema with
distinct groups of three children? How many times an individual child will go to cinema
before the group is repeated.

6 boys considered as 1 unit and 2 girls considered as 1 unit(orders are same)

Remaining boys =50-6-2=42
Each boy can be arranged in 1 way,so 42 boys in 42! ways
so,42!+1!+1!=44!
ans)44!

In the island of punycity license plates for cars consists of 3 letters.

2 license plates are considered if and only if they contain the same
3 letters in the same order .How many punycity license plates are
possible if the letter Q cannot occur as the last character and can
only occur when the letter immediately after it is U ?
a)17575 b)15625 c)15675 d)17650

we have 26 letters.
so without Q now we have 25 letters.
so the license consist of 3 letters.
so now no of combinations can be possible 25*25*25=15625.
now if u put q in the license then it must like this
Q U _ = for this combination we have 25 values
_ Q U = for this combination we have 25 values
now the total combinations will be=15625+25+25=15675. :)

In a hotel we can order two types of varities,but we can make 6 more variteis in
home.One can choose the
four varities with two from hotel as must.Find how many ways one can order.
a)14 b)15 c)56 d)28

in question there is 6 more varieties in home,not exactly 6 ,6 more than no of varieties can order in
hotel..so it will be (6+2)=8
and ans will be 1*8C2=28

There are 4 boxes colored red, yellow, green and blue. If 2 boxes are selected, how
many combinations are there for at least one green box or one red box to be selected?

lets fix that one box is green then no of selection of 1 box from red, yellow blue is 3C1......
lets fix that one box is red then no of selection of 1 box from green, yellow blue is 3C1......
so total combination=3c1+3c1=6
but red and green occur in both cases which are similar so we reduce solution by 1 so total number of
ways= 6-1= 5

In how many ways can the letters of english alphabets is arranged

so that there are 7 letters between the letters A & B and no letter is
repeated?

we select 7 letters from 26-2(exclude a,b)

so we select 24p7 ways
and we arranged in 2 ways(start with either a or b)
remaining 18 in 18! ways

A and B plays a game of dice between them. The consist of colors on their faces (
instead of numbers). When the dices are thrown, A wins if both show the same color.
Otherwise B wins. One die has 4 re face and 2 blue faces. How many red and blue faces
should the other die have if the both players have the same chances of winning?
a)3 red and 3 blue b) 2 red and remaining blue c) 6 red and 2 blue d) 4 red and 2 blue

Second dice has n red faces and (6-n) blue faces

For A and B to have an equal chance of winning probability of A winning = 0.5

Probability of A winning = [1/6 * (n/6)] + [5/6 * (6-n)/6] = 0.5
n/36 + (30-5n)/36 = 18/36
n + 30 - 5n = 18
30 - 4n = 18
4n = 12
n=3

The second dice must have 3 red faces and 3 blue faces

A father purchased dress for his 3 daughters. The dresses are of same color but diff size
and they are kept in dark room. what is probability that all the 3 will not choose their
own dress?

the arrangement will follow this way =>first girl will choose wrong dress AND second girl will wrong
dress AND the third girl will choose wrong dress.
since there are three dresses so .
=>first girl will choose wrong = 1-(1/3) = 2/3
now two dress are left
=>second girl will choose wrong out of two dress left => 1-(1/2) =1/2
now third girl will chose in only one way=1
so required probability= 2/3*1/2*1
=1/3 (ans)

Lady has 2 select gloves & hat from a basket. In the dark, she can distinguish hat &
gloves. 14 red, 20 blue, 18 green are there. Find probability that any selected glove
pair has same color.

no.of ways of getting 2 gloves(pair)form total(14+20+18=52) is 52c2=1326

no. of ways of getting 2 red gloves from 14 red gloves is 14c2=91.
no. of ways of getting 2 blue gloves from 20 blue gloves is 20c2=190.
no. of ways of getting 2 green gloves from 18 green gloves is 18c2=153.
so the total no of ways =91+190+153=434.
so the required probability is =434/1326=0.327 [answer]
A group of twelve persons is to be seated around a circular table. There are only two
women in the group find the probability that there is at least three men between the
two women. 1. 5/11 2. 7/12 3. 1/3 4. 2/9
5/11
total 3m+2w=>5
as it is a circular table it will be n-1
given n=12; therefore =(5)/(n-1); =5/(12-1); =5/11

If a traveller was offered 5 destinations in asia and 11 destinations in europe and asked
to chose utmost 3 destinations in asia and overall a 11 destinations. How many ways
can he travel

5c3*11c8+5c2*11c9+5c1*11c10+1=2256

A drawer has 4 red hats and 4 blue hats.find the probability of getting excatly 3 blue
hats when taking out 4 hats randomly out of the drawer and immediatly returning
every hat to the drawer before taking out the next?

possible
outcomes=(RRRR)(RRRB)(RRBR)(RBRR)(BRRR)(BBBR)(BBRB)(BRBB)(RBBB)(RRBB)(BBRR)(RBBR)(BR
RB)(RBRB)(BRBR)(BBBB)
so toatl 16
n we want exactly 3 blue hats
so 4/16=1/4 ans.

18 guests have to be seated,half on each side of a long table .4

particular guests desire to sit on one particular side and 3 others on
the other side .Determine the number of ways in which the seating
arrangement can be made

9p4*9p3*11!*2
4 guests one side out of 9 seats =9p4
3 guests another side out of 9 seats =9p3
remain 11 guest can sit in 11 seats any where =11!
the 3 and 4 perticular guest can change their side =2

In how many ways can 7 different objects be dived among 3 persons so that either one
or two of them do not get any object?
a. 381 b. 180 c. 36 d. 84

case 1: when one of them do not get any object, then the objects will be divided among 2 persons.
no. of ways to select 2 persons= 3c2.
no. of ways to divide 7 objects among 2 persons= 2^7.
so required no. of ways= 3c2(2^7-2) = 378 [when none of them get any object]

case 2: when objects are given to only one person.

reqd. no. of ways= 3c1 =3.
so, total no. of ways = 378+3 = 381

Paul the octopus who has been forecasting the outcome of FIFA world cup matches with
tremendous
accuracy has now been invited to predict ICC world cup matches in 2011. Wewill
assume that the world cup
contenders have been divided into 2 groups of 9 teams each. Each team in a group
plays the other teams in the
group. The top two teams from each group enter the semifinals ( after which the winner
is decided by
knockout).
However, Paul has a soft spot for India and when India plays any team, Paul always
backs India. Alas, his
predictions on matches involving India are right only 2 out of 3 times. In order to
qualify for the semi finals, it
is sufficient for India to win 7 of its group matches. What is the probability that India
will win the ICC world
cup?
a) (2/3)^10
b) (2/3)^9 + 8/3 * (2/3)^9
c) 8/3 * (2/3)^9
d) (2/3)^10 + 8/3*(2/3)^9

Solution is d)
Here winning prob is 2/3 ,so loosing prob is 1/3.

If India win all 10 matches then probability is (2/3)^10

If India loose any one match out of any 8 league matches then probability of winning 7 league
matches is (2/3)^7. and for one loose match probability is (1/3).
so these two is also arrange in 8 possible cases, which is arranged as 8{(2/3)^9 * (1/3)}.
Hence total probability is (2/3)^10 + 8/3 * (2/3)^9
How many different integers can be expresed as the sum of three distinct numbers
from the set{13,10,23,28,33,36,43,48}?

55 is rt ans..bcoz 8c3-1=55. one is subtracted bcoz (43+13+10=66) and (33+23+10=66)..

The telephone company wants to add an area code composed of 2

letters to every phone number. In order to do so, the company
chose a special sign language containing 124 different signs. If the
company used 122 of the signs fully and two remained unused, how
many additional area codes can be created if the company uses all
124 signs?

122p2=14762
124p2=15252
15252-14762=490 additional codes

A person has in is bag 14 notes of Rs. 10 each, 9 notes of Rs. 5

each, 4 notes of Rs. 2 each and 7 notes of Re 1 each. In how many
different ways can be contribute to a charitable find?

A. 599 B. 5999 C. 3528 D. 600 E. 2579

As per the questions,

14 notes of Rs10...so person can contribute Rs10 note in 14 ways.if person dont give Rs 10
note then number of ways of contribution are 14+1=15.
9 notes of Rs 5...so person can contribute Rs5 note in 9 ways.if person dont give Rs 10 note
then number of ways of contribution are 9+1=10.
and so on..
SO the number of ways can be contribute to a charitable fund 15*10*5*8-1=5999(option b)
answer
here -1 is used because person must have to use atleast one note.
Here SANDEEP is correct..

How many ways can Jason sit with his five friends in a row of six seats with an
aisle on either side of the row, if Jason insists on sitting one of the aisles?
a) 120 b) 240 c) 360 d) 540

ans is 240 : Because Jason must sit on either aisle, there are two places he can sit. And for
each of those two seats, there are five friends left for five seats, making 5! ways to arrange
them. So the answer is 2 * 5!, which is 240.
2 towns divided into 4 zones with 3 town in each. Every town of
same zone is connected to the town in its zone with 3 direct
telephone lines. The town from one zone is connected to town of
other zone by one direct line. How many direct lines are required?
Consider a town in Zone 1. There are 9 towns outside this zone.

Hence 9 lines are required to connect between this town and the other 9 towns.

Similarly, there are 12 towns, each with 9 towns outside its own zone.

So, totally, there will be 9*12 / 2 = 9* 6 = 54 lines inter zone (Divided by 2 comes here
because, after half, it is repeated).

Alternate way for inter zone:

1st zone each town has to connect to 9 other towns = 27 lines.

2nd zone – each town has to connect to 6 other zones, as the 1st zone is already accounted
for. So 6*3 = 18.

3rd zone – each town has to connect to 3 other zones, as the 1st and 2nd zone is already
accounted for. So 3 * 3 = 9.

4th zone is accounted for already.

Hence, totally 26 + 18 + 9 = 54 zones.

So, the total number of lines will be 36 (within zone) + 54 ( inter zone) = 90.

3 cars A, B & C are in the race. A is twice as likely to win as B and B

is thrice as likely to win as C. what is probability that B will win, if
only one can win the race ?
a) 1/2 b) 2/5 c) 3/10 d) 1/10

ANS= 3/10
LET C=X
B=3X
A=6X
THEN B'S PROBABILITY= 3X/(X+3X+6X)
B'S = 3/10

Thangam and Pandiyamma go for an interview for two vacancies.

The probability for the selection of Thangam is 1/3 and whereas the
probability for the selection of Pandiyamma is 1/5. What is the
probability that none of them are selected?
A. 3/5
B. 7/12
C. 8/15
D. 1/5

Answer: c
Probability of thangam being selected= 1/3 and him not being selected = 1-1/3=2/3
Probability of pandiyamma being selected=1/5 and him not being selected= 1-1/5=4/5
Probability that both are not selected = 2/3*4/5=8/15

There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are
drawn with replacement. What is the probability atleast three are
red!

ase 1 : 3 red balls : 6x6x6x15x15

case 2 : 4 red balls : 6x6x6x6x15
case 3 : 5 red balls : 6x6x6x6x6
favourable cases = cases ( 1 + 2 + 3) = 30456
total cases = 21x21x21x21x21 = 21^5
probability = 376/50421

if the probability of rain on any given day in pune city is 50% then
what is the probability that it rains on exactly 3 days in a 5 day
period?

probability it rains on 1st day=1/2

probability it rains on 2nd day=1/2
probability it rains on 3rd day=1/2
probability it rains on 4th day=1/2
probability it rains on 5th day=1/2
Probability of rain in any day of 5day period=1/32
probability that it rains on exactly 3 days in a 5 day period=5c3*(1/32)=10/32=5/16

Thirty days are in September, April, June and November. Some

months are of thirty one days. A month is chosen at random. Then
its probability of having exactly three days less than maximum of 31
is A. 15/16 B.1 C.3/48 D.None of these
Ans= 3/48

Sol: Prob(Getting Feb month of 28 days) = Prob(selecting non leap year) X Prob(Selecting
feb month after non leap year has chosen)

Prob(Getting Feb month of 28 days) = (3/4) x (1/12)

= 3/48
(Option C)

In his wardrobe, Pawan has 3 trousers. One of them is black, the

other is blue, and the third is brown. In the wardrobe, he also has 4
shirts. One of them is black and the other 3 are white. He opens his
wardrobe in the dark and picks out one shirt and trouser pair
without examining their color. What is the likelihood that neither the
shirt nor the trouser is black ?
Option
1) 1⁄12
2) 1⁄6
3) 1⁄4
4) 1⁄2
total 3 trousers in that,
no of black trousers =1
total shirts=4
no of black shirts=1
therefore,possibilty of picking trouser is black= 1/3
probability of choosen shirt is not black=1-1/3=2/3
possibilty of picking shirt is black=1/4
possibilty of picking shirt is not black=1-1/4=3/4
neither of them is black is =(2/3)*(3/4)=1/2
so answer is 1/2

There are two bags. One bag contains 4 white and 2 black balls.
Second bad contains 5 white and 4 black balls. 2 balls are
transferred from first bag to second bag. Then one ball is taken
from the second bag. The probability that the ball is white is ?

1) 42⁄165
2) 5⁄165
3) 48⁄165
4) 19⁄33

Ans = d
prof:
(prob of trans 2 white ball and prob of taking 1 white ball from 2nd bag) 0r
(prob of trans 2 black ball and prob of taking 1 white ball from 2nd bag) or
(prob of trans 1 white ball & 1 black ball and prob of taking 1 white ball from 2nd bag)
==> ((4c2 * 7c1) + (2c2 * 5c1) + (4C1 * 2c1 * 6c1)) / (6c2 * 11c1)
= 19/33

In a single throw with two dice, find the probability that their sum
is a multiple either of 3 or 4.
option
a) 1/3
b) 1/2
c) 5/9
d)17/36

5/9
20 cases which lead sum divisible either by 3 or 4 and total no. of cases is 36

A question has 7 options. If he chooses correct answer he can get

one mark. If he chooses wrong option he looses one mark. If he
chooses randomly he gets zero marks. Any way he identifies 2
options and eliminates them. If he chooses randomly how many
marks can he gain?

probability to choose correct option=1/7 and associated marks is 1;

probability to choose incorrect option=6/7 and associated marks is x(let);
total marks={(1/7)*1}+{(6/7)*x}=0, i.e. x=-1/6;
now 2 options are eliminated,
so,now,
{(1/5)*1}+{(4/5)*(-1/6)}=1/15;
ans is 1/15.

Of 38 people in my office, 10 like to drink chocolate,15 are cricket fans and 20 neither
like chocolate nor like cricket, how many people like both cricket and chocolate?

a. 7
b. 10
c. 15
d. 18

20 neither like chocolate nor cricket

therefore,people likes either chocolate or cricket are+38-20=18
so, now pele like both=people like chocolate + peole likes cricke -
people like either chocolate or cricket
=10+15-18
=7
so,option a) is the correct ans.

Out of 800 families with 4 children each, how many families would
you expect to have 2 boys and 2 girls in each family?

500
400
100
200
300

for this type of problem,always remembr d follwng patterns:The sample space for a family having four
children is:

{(BBBB), (GGGG), (BGGG), (GBGG), (GGBG), (GGGB), (BBGG), (BGBG), (BGGB), (GBGB), (GBBG),
(GGBB), (BBBG), (BBGB), (BGBB), (GBBB)}

B - stands for Boy and G - stands for girl

a) For a family to have two boys and two girls it is: {(BBGG), (BGBG), (BGGB), (GBGB), (GBBG),
(GGBB)} = 6 elements
Its probability = 6/16 = 3/8 = 0.375

Hence, the probability for 800 families = (0.375) x(800) = 300

b) A family to have atleast one boy = 1 - P(All gilrs) = 1 - 1/16 = 15/16 = 0.9375

Hence, the probability for 800 families = (0.9375) x (800) = 750

c) Probability of children of both sexes = 1 - {P(all boys) + P(all girls)}

= 1 - 1/8 = 7/8 = 0.875

Hence, the probability for 800 families = (0.875) x(800) = 700

Three cars A, B and C are participating in a race.A is twice as likely

as B to win and B is thrice as likely as C to win. What is the
probability that B will win, if only one of them can win the race?
p(A)=2*p(B)
P(B)=3*p(c)
P(A) +P(B)+P(C) =1
2 * P(B) + P(B)+P(B)/3 =1
10*p(B) /3=1
P(B)=3/10

Aniket and kumari are a married couple (dont ask me who he is and
who she is)! They have two kids, one of them is a girl. Assume
safely that the probability of each gender is 1/2.
What is the probability that the other kid is also a girl?
Hint: It is not 1/2 as you would first think.

The following are possible combinations of two children that form a sample space in any earthly
family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample
space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

A and B throws dice.what is the probability that A's throws is not

greater than B's throws?

A's throw throw could be greater than B by following ways

(2,1)
(3,2) (3,1)
(4,3) (4,2) (4,1)
(5,4) (5,3) (5,2) (5,1)
(6,5) (6,4), (6,3) (6,2), (6,1)
So in total, there are 15 possible outcomes in which the throw could exceed.

So in the probability in which A's throw can exceed is

15/36 or 5/12

Probability of failure = 1-p

= 1-15/36 = 21/36 = 7/12

probability that A's throws is not greater than B's throws=7/12(ans)

a drawer holds 4 red hats and for blue hats. What is the probability
of getting exactly 3 red hats or exactly 3 blue hats when taking out
4 hats at random and return every hat to drawer before taking out
another one

It is quite simple.

When you take out hats one by one after replacement, there are equal chances of getting red or blue
hat.

So possible outcomes are

RRRB, RRBR,RBRR,BRRR, BBBR,BBRB,BRBB,RBBB.. with exactly three red or blue hats.
RRBB,BBRR,RBRB,BRBR,BRRB,RBBR, BBBB,RRRR.. with other combinations.

so out of 16 possible combinations , eight are desired combinations.

so probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats at random
and return every hat to drawer before taking out another one = 8/16 =1/2

there are two boxes one containing 10 red balls and other
containing 10 green balls.you are allowed to move the balls
between the boxes so that when you choose a box at random an a
ball at random from that box the probability of getting a ball is
maximised.find the maximum probability.
1)14/19,
2)3/4,
3)1/2,
4)37/38
To maximise probability, transfer all red balls except 1 from box1 to box2.So now box1 will have 1 red
ball and box2 will have 19(9red + 10 green)balls. Now probability of selecting box1 from those is 1/2.
Since there is 1 red ball only, prbability of selecting red ball will be 1. So for box1 total is
(1/2)(1)=1/2. Similarly for box2 its (1/2)(9/19). Add both you get 14/19

A Box has 13 white chips, 7 blue chips and 6 green chips. What is
the probability that, if two chips are drawn from the box in
succession, one is blue and other is white.
option
a) 7/50
b) 20/26
c) 4/20

7/26*13/25=7/50 ans...

here is magic dragon with 3 heads & 3 tails....u have a magic sword
tht can cut 1 head,2 heads , 1 tail & 2 tails...if 1 head is cut another
head grows...if 1 tail is cut 2 new tails grow...2 tails are cut a new
head grows...& if 2 heads r cut nothing grows...find minimium
number of swings 2 cut all heads & tails

1 cut two tail=4 head 1 tail

2 cut 1 tail=4 head 2 tail
3cut 1 tail= 4 head 3 tail
4 cut 1 tail=4 head 4 tail
5 cut 2 tail= 5 head 2 tail
6 cut 2 tail=6 head 0 tail
7,8,9 cut 2 heads one by one.
therefore ladies and gentleman our ans is 9

The bacteria has the probability of split into 3 and probability to die
is 1/3rd of the total bacteria.Let the probability is P.Some of them
survived with probability 1/5.Then which among the following
relation is true?
a)P=1/3+1/5*3
b)P=1/5*(1/8-3)

here prob for dying = 1/3

hence prob of survival = 1/5
if they survive, they will be splited into 3 more bacteria
hence, p = (1/3) + (3* 1/5)
A gambler has in his pocket a fair coin and a two-headed coin. He
withdraws a coin at random and
flips it twice. If the first flip is Heads, what is the probability that the
second flip is Heads ?

probability of finding head with fair coin=1/2*1/2=1/4.

probability of finding head with unfair coin is 1/2*1=1/2.
total probabilty of getting head=1/4+1/2=3/4.

Paul the octopus who has been forecasting the outcome of FIFA
world cup matches with tremendous accuracy has now been invited
to predict ICC world cup matches in 2011. We will assume that the
world cup contenders have been divided into 2 groups of 9 teams
each. Each team in a group plays the other teams in the group. The
top two teams from each group enter the semi finals (after which
the winner is decided by knockout). However, Paul has a soft spot
for India and when India plays any team, Paul always backs India.
Alas, his predictions on matches involving India are right only 2 out
of 3 times. In order to qualify for the semi finals, it is sufficient for
India to win 7 of its group matches. What is the probability that
India will win the ICC world cup?

ans is : (2/3)^10 + (8/3)*(2/3)^9

given that prob. of India winning the match is 2/3.

taking that India wins the semi-final and final.Prob is (2/3)^2

this gets multiplied to the league stage...

in league stage, if India has won all the games, the prob. of winning is (2/3)^10 for the 10 matches
played...

if India had lost anyone of the matches, the prob. is

(8/3)*(2/3)^9..
where 8(1/3) is the prob. of losing anyone of the 8 league matches.

The crew of a rowing team of 8 members is to be chosen from 12

men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that
there are two rows, each row occupying one the 2 sides of the boat
and that each side must have 4 members including at least one
women. Further it is also known W1 and M7 must be selected for
one of its sides while M2, M3 and M10 must be selected for other
side. What is the number of ways in which rowing team can be
arranged.

7C1*14C2*4!*4!
second side will be definitely contain woman as there should be one women on each side so no of
ways 7C1
and remaing 2 places we have to fill from 14 remaining people ways 14C2 and each side can be
arranged in 4! Ways

There are 14 spots. Each spot has 8 seats. 28 people seated in all
spots. No similar number of people sat in any spot. How many spot
left with no people at all.
a) 7
b) 10
c) 8
d) 12

ans: 7

total 14 spots with 8 chairs

at each spot different number of people
so, possible combinations
1+2+3+4+5+6+7+8
but,
only 28 people their
so,
1+2+3+4+5+6+7=28
so 7 possibilities
vacant=14-7=7

There are two boxes, one containing 21 red balls and the other
containing 25 green balls. You are allowed to move the balls
between the boxes so that when you choose a box at random and a
ball at random from the chosen box, the probability of getting a red
is maximized. This maximum probability is
(a) 0.5 (b) 0.63 (c) 0.72 (d) 0.48
0.72
we will keep 1 red ball in 1 box and rest 20 in another box.
then total prob.=
1/2 +1/2*2/9=13/18

A bag contains 1100 tickets numbered 1, 2, 3, ... 1100. If a ticket is

drawn out of it at random, what is the probability that the ticket
drawn has the digit 2 appearing on it?

A. 291/1100
B. 292/1100
C. 290/1100
D. 301/1100

within 1 to 100 there are 19 numbers containing digit 2.2 first is 20 to 29 there are 10 numbers and 2
12,32,-,- ------92.
and 200 to 299 there are 100 numbers. so 1100 contain 290 numbers containing digit 2.(100---19 so
100 contain 190 ,excluding 200 to 299) 190+100=290.
so 290/1100 is answer.

4 men throw a die each simultaneously. Find the probability that at

least 2 people get the same number.

13/18 as 1-(probability of not getting same no.)=1-((6*5*4*3)/(6*6*6*6))

The first person rolls a die, and gets a number.

The second person has a 5/6 chance of rolling a different number.

The third person has a 4/6 chance of rolling a number different from those two.

The fourth person has a 3/6 chance of rolling a number different from all three.

The odds that no two will roll the same number is 5/6*4/6*3/6 = 5/18 = .2777...

That's the odds that no two will roll the same number, so the odds that two (or more) of them will roll
the same number is 1-5/18 = 13/18 = .7222...

100 coins are tossed. Probability of getting 50 heads is equal to

probability of getting 51 heads. That probability is??
p=probability of getting one head in one trail
the total number of trails n=100
then by using binomial dist with parameters n,p
P(50)=p(51)
((1-p)/50)=(p/51)
p=(51/101)

a bag contains 2/3rd blue balls and rest are pink.then there are
some part of blue balls are defective and some part of black balls
are defective(i don't recall the exact value)n tatal non defective
balls are 146.then total no. of balls in bag.?
ANS:432

a bag contains 2/3rd blue balls and rest are pink.5/9 blue balls are defective and 7/8 pink balls are
defective.total non defective balls are 146.then total no. of balls in bag.?

sol:let balls in bag be x

blue=2x/3 ,pink=x/3
non defective blue balls=(4/9)*(2x/3)=8x/27,
....do........ pink =(1/8)*(x/3)=x/24
so (8x/27)+(x/24)=146
on solving x=432.So total balls=432

What will be the no. of different ways to answer six questions

having 3 different options (i.e. yes/no/Don't know) ?
a.6C3 b.6p3 c.6!*3! d.9!

A bag contains 3 balls of 11 different colors each. Find the min no of

chances to find at least 3 balls of same color?

Two percent of the parts produced by a machine are defective. Twenty parts
are selected at random. Use the binomial probability tables to answer the
following questions.
a. What is the probability that exactly 3 parts will be defective

Argentina had football team of 22 player of which captain is from Brazilian team and
goalkeeper from European team. For remainig players, they have picked 6 from
argentinan and 14 from european. Now for a team of 11 they must have goalkeeper
and captain so out of 9 now, they plan to select 3 from argentinian and 6 from
European. Find out no. of possibilities available for it.

The Israelis MOSAD group has n Assassin for the role of Killing ISIS
Terrorists, who help in smooth killing of Terrorists who intruded in
their country. Every Agents can either manage to shoot a terrorist
or miss it.

Out of these n Killing Assassins, m assassins try to shoot the

terrorist fairly, that is, they are unbiased, while the rest of them are
biased. The probability that a terrorist will be hit by a shot if the
Killing Assissin is biased is 2/3, otherwise the probability is 1/2.

Now MOSAD Chief figures out this mess by selecting a Killing

Assassin randomly. He tricks that particular Killing Assassin in
shooting the same terrorist twice. The first time, the Killing Assassin
shoots the terrorist, while the other time, he ends up missing it.

Chief wants to know what are the chances that the selected Killing
Assistant was NOT biased.

Given data is Total number of assassin is 8 and out of 8, 5 are

unbiased assassins.

Answers may be one of the option

1) 15/23 2) 17/49 3) 1/1 4) 14/19

A man has 7 friend among them 4 are female and 3 are male. And his wife has 7 friend
among them 4 are male and 3 are female. 6 person were invited for party. What is the
probability that there were 3 female, 3 male, 3 female friends of man and 3 male
friends of his wife?