Ch ae Probability 3. {(h) Number of sample space = 6 * 6 = 36 ‘Sum is 10 oF 11 = (4,6), (5, $) (5, 6). (6,4), (6, 5) 5 6 (@) ax? + hx += 0; For equal roots, 6? = 4ac Required probability b @o Total a 0 z ay 1 3 5 ° 4 [aaanes [3 5 : 0 6 G3) 1 Required probily = (@) Total number ways we can choose three integers from 20 integers = °C, (1) We will get all three integers are multiple of three (3,6, 9, 12, 15, 18) = °C, (1) We will get sum divisible by 3, if all three numbers when divided by 3 givens remainder of I (1,4, 7, 10, 13, 16, 19) = 7C, (Us) We will get sum divisible by 3, ifall three numbers are when divided by 3, gives a remainder of 2 (2,5, 8,11, 14,17, 20)=7C, (IV) We will get sum divisible by 3, if one of them is divisible by 3, one of them when divided by 3 gives a remainder of 1, one of them when divided by 3, gives a remainder 227, *7C,*6C, > $QH7G+7G+71Gx7Gx%C,_32 TG 8S (@) Total number of arrangements of 8 teachers and 4 students in circular table = (8 + 4 ~1)!= 11! Number of arrangements of 8 teachers and 4 students in Circular table such that no two students sit together = 8Cy-41(8- I) =8Cq-41-71 Probability +, Required probability 5. (b) Total words can be made out of SENSELESSNESS' 131 raat 180180; Total words can be made with all the Es togetier = "3-22 2520 iy = 2820, 2 4. Required probability = 2520. 2, Required probability = Feo Te5 = F453 () Total number of ways in which two numbers can be selected from (1,2, 3, 10) = 10? be yis divisible by 6 = (6-y) («+ ys divisible by 6 = cither (xy) is divisible by 6 or (x + y)is divisible by 6. So, possible outcomes are Values of | Values fy | No, of eases Hi 157 3 2 | 24s] 4 3 39 2 4 [argo] 4 s 5.7 3 6 6 1 7 As 3 8 [aaa] 4 9 93 2 to | imsas] 4 30 Required probability = =% 10 (©) Let A be the event that a multiple of 3 does not appear. 10 4.2 P(A) = &= 3% Now, Probability that a multiple of 3 ae does not appear on any dice = G (2) Total numberof five digits htinber = 4.4.3.2.1 = 96 For number divisible by 4 contain Last wo digit 04, 40 and 2 Last two digit 12, 32 and 2. Total number divisible by 4 = 30 5 Required probability = 2 +24328 9. 10. 2. 1. 4 () Since, otal number of ways of distributing of 7 balls in 4 box = 47; Total number of ways such that first box contain 3 balls = 7c, x3* 7eyx3* 35/3)" 64\4) ° x34 So, required probability = ‘ar (©) “> We know, if two numbers x and y are choosen at random without replacement from the set (1,2,3, 4, 5K), then the probability that x4 — y* is divisible by 5 is 1K-5 : Given set= {1,2,3,4,5) > K=1 Wei-s 12 SSxI=1) 5x4 5 55K=D So, required probability = (a) Given numbers 2,3, 5,7, 11, 13 Total no. of ways 3 chits can be choosen = *C, ‘Now, we observe the outcomes for sum divisible by 3. So, the pairs whose sum divisible by 3 are (2, 3, 7), 2, 3,13), 2 5, 1D, G5, B.S, 13), 8, 7, 1), G, 10, 13) Hence, required probabilit (b) Case 1 : 2leters are repeating, No. of such ways= 1 Case 2: 2 letters are repeating and 2 are non repeating No. of such ways = 2% 8C, = 56 Case 3: All letters are IRB 7X6 f such ways = °C, No, ofsuch ways =9Cy = 2 126 1456 57. (1+56+126) 183° 61 (b) Total number of possible three numbers = “°C, = 19,600. Now a, 6, care in AP. => 2b =a +e=even Therefore a care either both even or both odd ‘There are 25 odd and 25 even numbers in {1,., 50) Favourable case = *5C, + ¥C, = 600 3 19600 98 (a) Let E,, E,, Ey, E, be the events that boy wrote the words ABILITY, PROBABILITY, FACILITY, MOBILITY respectively. 19 Required probability = Probability = L So, P(E,) = P(E,)= P(E,) = P(E.) = 3 Abe the event that “LI’ left after erasing 15. 16. 1. 18, Des Using, Baye’s theorem 5 1%) T,1,1 TT rw) a 16 (a) Number of possible outcomes of event when 2 dice are thrown are 6* = 36 Outcomes in which number are not coprimes = (2,2).2,4),2.6),3, 3), G6). (4,2), (4,4), (4,6). (5, 5)s (6, 2) (6,3) (6.4), (6, 6)} Favourable outcomes (getting coprimes number) ~ (Total Not coprimes number) = 36 ~ 13 = 23 3 6 (©) Total card = 52, No. of suit=4 ‘Card having composite number in each suit =4,6,8,9, 10 Card having multiple of 3 in each suit=3, 6,9 ‘No. of cards having composite number in 4 suits. (4,8, 10) = 3 x 4= 12 and No. of cards having multiple of 3 in all suits (only 3) = 1x 4=4 No, of cards having number (6,9) in all suits =2.* 4= 8 Required probability = (126, x44) +12(8C} 124) + (BC, x4C}) + (B8CQ) 7 52C, Required probability = 48496432428 102 320 663 (@) Number of way to matched 5 men with wives =5-4-3-2-1= 120; Numberofwaynomanismatchedwith none ofhis wife = Ds 44 Probability = 55-35 (a) Sum of numbers can be 2, 3, 5,7, 11. Multiples of 3 are 3 and 6, Sum of numbers such that 3 or 6 appears on top among the pair of numbers can be 5, 7, 11. 5 (2,3)(G, 2pand 11 > (5,6) (6,5) 7 4,3),4)(6,1)(1,6) No. of eases = 8; Sum can be prime in following cases (yD, 2 Us 2), 2, 3s Bs 2s 4 3), B42 (6, Dy (1, 8), (5, 6). (6,5). (4, 1s (Ls (5, 23,2, 5) 8 No. of cases = 15; Required probability =ERTELNy 19. 20. 21. m4, 25, 26, () Prime numbers among same suit are 2,3, 5,7 Their number = 4; No. of face cards = 3 Total no. of cases = 13 « 12; Favourable cases = 4% 3 + m4 2 in (2) A noc-leap year contains 365 days and there are {otal $2 weeks and 1 day ina non-leap year. Hence there should be $2 Sunday also, But the 1 extra day apart from those 52 weeks may be either Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or a Sunday. Cleatly, total possible events are = 7 and Number of 3 «4=24; Required probability favourable events = 1; Hence probability = 4 (b) Total no. of balls = 10 ‘No. of blue balls = 4; No, of red balls = 6 No. of ways to draw three red balls = 6¢, Soy 1 10¢, 6 «+ The required probability i (©) AandB are Event that A, B speaks the truth 4 P(A)= 2 P(A*)=4;P(B)=3=99| Gi) Now, th8 requited probability ~*P(A) . P(BS)"+ P(AS) 4113-7 ahd l3 5°4°5 "4" 20 (a) Let the probability of A, B, C winning are a, b, ¢ respectively. Then,a+b+e=1 @ b Given that, a 2b &b= Ic; s.c= > rams 2i-b+ =i bn 42 So, the probability of horses A, B & C to win are 7" respectively. (b) If £, ¢ Ey Then P(E,) s PE) «+ Statement (iv) is not satisfied by P. (a) S= ((T, 1), (HH, (7, 1), (HD) 3 Probability of not getting tal 4 a) +. The required probability = (3) (©) When only event A occurs, then it can be written as (An 8°), And when only event B occurs, then it can be written as (A° mB). Then, ‘exactly one of the events 4, B to oceur can be written as : (AMBC)U(ACB) 2. 28, 2». 30. 3. 32, 33. 35. 4329 (2) Number of sample space =2«2*2=8 (HHT. HTH, HTT, THH, THT, 1TH one 874 (©) n(A)= members proficient in mathematics = 19 ‘n(B) = members proficient in statistics = 16 (AWB) = n()+0(B)—n(AB) => 25=19416—n(ANB) => n(ArB)=10 (AB) 5 (b) Total no. of ways of selecting 2 students out of 100 = 109, and No. of ways in which both of us enter the same section = °C, + °C, 20 +c We (0) = P(4UB)=05+04-03=06 So probability neither A nor B= 1 - 0.6 = 0.4 (@) Twoset d and B can be selected in 2" ways. The no. of ways of selecting two sets such that their union is S and intersection is 2%. Now, favourable outcome = { So required probability Now, P(AMB) = +. Requited probability yaa Therefore the probabiliy= 35 oy = Sy (8) No, of events in sample space when a pai of dice is rolled 24 times = (6 « 6)°* = (36). ‘No. of event of not getting on both of the dice = 36 a5 Go @) #+ar+b=0 > D20>a-4520 +. The required probability is, a=3, b=1,2 b=1,2,3,4 b= 1,234 Possible pairs = 10'=> Total pairs = 12> Probability: 3 (a) Total floors = 7, Total number of ways to exit = 7 IFS persons leave at 5 different floors then number of ‘ways = arranging 5 people in 7 ways =P, 360 2401 Ps__7 Required probability (©) Total number of elements in sample space = 6% 6x 6=216 and Ways to get different sumbers on all 3 dice =6* 5 4= 120 120 Probability = 37g 54330 36. 37, 39. 40. i. a. 4. (@) P(AUB)=P(ANB) + P(AUB)=P(A)+P(B)~P(AMB) . 2P(ANB)=2P(AUB) = P(A)+ P(B) Clearly P(A) +P(B)= 2P(A-.B) wich cart be aways 1. +. P(A) + P(B) = 1 is wrong (©) ln non-leap year, there are 365 days i.e. 52 complete ‘weeks and I extra day. That day can be any of 7 days. P (53 Sunday) = P (Last | day Sunday) 4 (53 Thursday) = ; Probability = (4) P[AI of three bottles ate of same variety] 654 Hed 321 25 286 13.100 34 286 286 (b) P card drawn is a face card (@) PIAII four cards are ofthe same suit] Be, +13c, +13e, +13e, 44 5c, 4165 (d) Number of slip for II year student = 3 * 100 = 300 Number of slip for Il year student ‘Number of slip for I year student 300 (@) Probability of being two kings is 4 Cd P(k)= get Wend 1 10 (©) P(A) == and P(B) = io How probability of solving the problem can given as P(A), P(A) P(B)oeP(X)P(B)orP(AyPCB) O31 77 10 “10 10 “10 "i0 Ox7 10x10 44, 45, 46. 47. 48, 19. (©) The bag contains W =3, Hence P(R) }+P(A) P(B) Px Similarly P (y fail) = P (Y" 2 33 Probability of atleast 2 passing = (Any 2 pass and | fail) + (AIL 3 pass) = P (X).P (Y).P (Z') + P(X) PCY’) P(Z) + P(X). P(Y) P(2)* P(X). PCY). PZ) o(bebd}e(bebad) (Sebel a(etad) ot SS) US aS) Sa SS) 6 (b) Product of 2 numbers is even if cither both are even or Lis even and | odd. Number of ordered pair: for 1 even and | odd = (Selecting 1 odd out of 3 and selecting 1 even out of 3) * 2 (a, b) and (b, a) are difference hence multiplied by 2 =283C, 3C, = 18 and both even: (2,2) (2.4), (2,6) (4, 2.4.4.4. (6,2), (6.9. (6.6) Total numberof favourable outcome = 18 +9 = 27 ‘Total outcomes = 36, Probability = 27/36 = 3/4 (d) Favourable outcomes => One ace and one queen and. | jack. There are 4 cards of each selecting | out of 4 can be done in 4C, x 4C, x 4C, = 4x 4x4 ways Total 3 outcomes are selecting out of $2 °C, = 22100 dxdxd 16° 22100 * 3525 (a) Total number of 5 digits numbers by using 1, 2, 3, 4, 5 without repetition = St Now, For making numbers divisible by 4, the last 2 digit number must be divisible by 4. This is possible in following cases: (a) ending with 12 ; Rest 3 place can be filled by 3! ways (b) ending with 24: Rest 3 place in 3! ways = 6 numbers (©) ending with 32: 3 places in 3! ways = 6 numbers (d) ending with 54: 3 places in 3! ways = 6 numbers ‘Total number of numbers divisible by + =6+6+6+6=24 PZ pass)=P(Z)= Required probability = :. Required probability = 2. a - t$0, (€) (Prive tickets) ~ 10, n(No prize ticket) = 25 2s P(No prize| 2 (No ps Gs $1. (@) Product of 2 nambers willbe odd only if both are edd Posy is even) = 1 - Play is odd) 1 Px odd) * Poy odd) = 52. (a) Pfking and spade card] = Pfking of spade and other spade card) + P(king spade ‘other than spade and spade card) -f1.2,3,3).4 . eaten) s (a) Favourable cases = ((2, 4), (2,6), 3.4). (3, 64 (5,4), ©), (4,2), (6,2), (4,3) (6, 3) (4, 9), (6, 5)} 2, Total no of cases = 6°=36 Probability 363 54. (a) Let A be event of selecting black ball 2 Probability of selecting black bal fom umn 1== Probability of selecting black al fom um 2=2 Since, one um is seleted at random hence Selecting one ura have probability 172 2) 1/29) » 2 70 (b) One card is selected at random fiom 27 cards numbered from |t027 => n=27 Possible numbers which ae even or divisible by 5 A= (2,4,5,6,8, 10,12 1415, 16, 18, 20,2, 24, 25,26 = 16 :. Probability = re 56, (b) Probability of drawing 3 white and 6 black balls 2 Gx7C, _ 1x10 7 9 Ro, 12-10 22 51, (8) Probability of speaking truth by 4 and B. 4 3 PUA) = and PUB) = ‘They will contradict each other in two case, Case: A > Truc, B+ False Case I: A False, B > True 58. 59. 60, 66. aga Probability = P(A) P(A") + P(B) P(A’ O} @ (Bonus) The perfect cube numbers in given set (1, 2, 3, 4 vy 1000) are 1, 8, 27, 64, 125, 216, 343, $12, 729, 1000 and the natural numbers having odd number of divisiors are perfect square numbers and the perfect square numbers in given set are I, 4,9, 16,25, 36,49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961 0431 1000 Hence, required probability = ‘1000 500 | (b) There are 50 numbers divisible by 2 in set {l,. wy 100}; Let E = {2, 4, 6, 100} ‘and the numbers in set E are divisible by 3 or S are {6, 10, 12, 18, 20, 24, 30, 36, 40, 42, 48, 50, 54, 60, 66, 70, 72, 78, 80, 84, 90, 96, 100} sored pays «2 (@) Since, D <0 for equation? +x+ 1 Since, o,@ areroots of 32 +x+ 1 = x otal tor ‘Total numbers of ways for (a,b, c)=6 x6 6 ‘Now, suitable values for (a, b, c)=6 x 4x2 6x4x2_ 8 2 2. Required probably = PO = eae 0, Hence let o (©) Let minimum percentage ofthe population ofthe town that should fall sick in order to raise the threat level is x. ‘Now, according to the question 10% of x= 5 or x= 50% (b) Ifa die is thrown tree times, Total possible outcomes, n = 6 = 216 For getting a larger number on its face than the previous ‘number each. Then, favourable outcomes = *C, = 20 rms " ; itity= 2 = 20% low required probabily= = 37 = zy (A) Let $= (2k|-9 P(4) = B. 3 «16 (a) Ifball is neither red nor green then it must be blue. m. ‘Number of Blue Balls P@blne bal) = Tea number of balls ie ee “B4746 213 @® 7% @ (@) Since total distribution = 312 Number of distribution when first box contains three 15. n. balls = "7C, x 2B. @) 79. (a) Let E be the event that A and B met at a party, Fbe the event that A and B met at a sports club and D be the event that 4 and B will dine together, =4 peeyer-4-5, of 2) 22 p(2)_1 nodes 2) 21(2) 9 roy=0e(2) rey 2(2) = ‘The probability that they disperse without line together. P(D')=1-P(D)