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The document contains two probability word problems. The first problem asks for the probability that three marksman simultaneously hitting a spinning spherical target will hit points on the same hemisphere. The solution shows that the probability is 1, since any two points define a hemisphere and the third point can fall in either hemisphere. The second problem asks which of three dice rolling events is most likely: rolling at least 1 six with 6 dice, at least 2 sixes with 12 dice, or at least 3 sixes with 18 dice. The solution calculates the probabilities and determines that rolling at least 1 six with 6 dice is most likely.

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Justine Marowe Austria
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252 views6 pages

Pstat

The document contains two probability word problems. The first problem asks for the probability that three marksman simultaneously hitting a spinning spherical target will hit points on the same hemisphere. The solution shows that the probability is 1, since any two points define a hemisphere and the third point can fall in either hemisphere. The second problem asks which of three dice rolling events is most likely: rolling at least 1 six with 6 dice, at least 2 sixes with 12 dice, or at least 3 sixes with 18 dice. The solution calculates the probabilities and determines that rolling at least 1 six with 6 dice is most likely.

Uploaded by

Justine Marowe Austria
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Three marksman simultaneously shoot and hit a rapidly spinning spherical target.

What is the probability that the three points of impact lie on the same hemisphere?

A. 0 C. 1

B. 1/2 D. 2/3

Solution:

A great circle can always be drawn through any two distinct points on the surface of
the sphere.

Say the marksmen will hit the points ​P​1​, ​P2​ ​, and ​P3​ ​. Draw a great circle through​P1​
and ​P​2​ dividing the sphere into two hemispheres. Since ​P1​ ​ and ​P​2​ are on the
boundary of these hemispheres, they belong to both hemispheres. Now ​P​3​ can be on
any of the two hemispheres. Hence, the three points will always be on the same
hemisphere.

Probability, ​P​ = 1.0

Samuel Pepys wrote Isaac Newton to ask which of three events is more likely: that a
person get (​a​) at least 1 six when 6 dice are rolled (​b)​ at least two sixes when 12 dice
are rolled, or (​c​) at least 3 sixes when 18 dice are rolled. What is the answer?

A. (​a)​ is more likely than (​b​) and (​c​)


B. (​b​) is more likely than (​a​) and (​c)​
C. (​c​) is more likely than (​a​) and (​b​)
D. (​a)​ , (​b​), and (​c​) are equally likely

Solution:

Event (​a​) In rolling 6 dice:


Probability that no six will come out
Q=(5/6)​6​=0.335​Q=(5/6)6=0.335
Probability that at least 1 six will come out

P​a​=1−Q​Pa=1−Q

P​a​=1−0.335=0.665​Pa=1−0.335=0.665

Event (​b)​ In rolling 12 dice:

Probability that no six will come out


Q=(5/6)​12​=0.112​Q=(5/6)12=0.112

Probability that at least 1 six will come out

Q​1​=1−0.112=0.888​Q1=1−0.112=0.888

Probability that exactly 1 six will come out

Q​2​=(1/6)​1​(5/6)​11​×12=0.269​Q2=(1/6)1(5/6)11×12=0.269

Probability that at least 2 sixes will come out

P​b​=Q​1​−Q​2​Pb=Q1−Q2

P​b​=0.888−0.269=0.619​Pb=0.888−0.269=0.619

Event (​c​) In rolling 18 dice:

Probability that no six will come out


Q=(5/6)​18​=0.038​Q=(5/6)18=0.038

Probability that at least 1 six will come out

Q​1​=1−0.038=0.962​Q1=1−0.038=0.962

Probability that exactly 1 six will come out


Q​2​=(1/6)​1​(5/6)​17​×18=0.135​Q2=(1/6)1(5/6)17×18=0.135

Probability that exactly 2 sixes will come out

Q​3​=(1/6)​2​(5/6)​16​×​18​C​2​=0.230​Q3=(1/6)2(5/6)16×18C2=0.230

Probability that at least 3 sixes will come out

P​c​=Q​1​−Q​2​−Q​3​Pc=Q1−Q2−Q3

P​c​=0.962−0.135−0.230=0.597​Pc=0.962−0.135−0.230=0.597

P​a​ > ​P​b​ > ​P​c

The most likely to happen is (​a)​ ← Answer: ​[ A ]

From where he stands, one step toward the cliff would send the drunken man over
the edge. He takes random steps, either toward or away from the cliff. At any step his
probability of taking a step away is 2/3, of a step toward the cliff 1/3. What is his
chance of escaping the cliff?

A. 2/27 C. 4/27

B. 107/243 D. 1/2

Solution

Formula:
P=1−(q/p)​^n

Where

P​ = probability of escaping the cliff


n​ = number of steps away from the cliff = 1
q​ = probability of taking a step toward the cliff = 1/3
p​ = probability of taking a step away from the cliff = 2/3
P=1−(1/32/3)​1​P=1−(1/32/3)1

P=​12​P=12 ← Answer: ​[ D ]

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