8/9/2019 PS1Answer

1/2

Probability and Statistics IISER Pune January 2015

Assignment 1, Page 1 of 2 January 2015

1. Amit throws six dice and wins if he scores at least one ace. Basu throws twelve diceand wins if he scores at least two aces. Who has the greater probability to win?

Answer: Prob (Amit loses)= 56

66; Prob(Amit wins) =0.665.

P(Basu loses)=12511

612+ 5

12

612; Prob(Basu wins)=0.618.

2. In a certain colony, 60% of the families won a car, 30% own a house and 20% wonboth a car and a house. If a family is randomly selected, what is the probability thatthis family owns a car or a house but not both.

Answer: C=car, H=house. P(C) = 0.6, P(H) = 0.3, P(CH) = 0.2.P(C CH) + P(H CH) = (0.6 0.2) + (0.3 0.2) = 0.5.

3. Given thirty people, find the probability that among the twelve months there are sixcontaining two birthdays and six containing three.

Answer:30!

2666(126 )

1230 .

4. A coin is tossed until for the first time the same result appears twice in succession.Describe the sample space. Find the probability of the following events: (a) theexperiment ends before the sixth toss, (b) an even number of tosses is required.

Answer: Sample Space= {H H , T T , H T T , T H H , H T H H , T H T T , . . .}.P(ntosses required) = P(n) = 1

2n1.

(a) P( the experiment ends before the sixth toss) = P(2) + P(3) + P(4) + P(5) =12+

122 +

123 +

124 =

1516 .

(b) 23

.

5. Ifn distinct balls are placed at random inton distinct cells, find the probability thatexactly one cell remains empty.

Answer: (n

2)n!nn

6. Consider the 24 possible arrangements (permutations) of the symbols 1234 and attachto each probability 1

24. Let Ai be the event that the digit i appears at its natural

place (where i= 1, 2, 3, 4). Verify that P(Ai Aj) =P(Ai) + P(Aj) P(Ai Aj).

Answer: P(Ai) = 3!4! =

14 . P(Ai Aj) =

2!4! =

112 . Thus P(Ai Aj) =

14 +

14

112 =

512

7. A closet contains n pairs of shoes. If 2r shoes are chosen at random (with 2r < n),what is the probability that there will be (a) no complete pair, (b) exactly onecomplete pair, (c) exactly two complete pairs among them?

Answer: (a) (n

2r)22r

(2n2r

) , (b)

(n1)(n1

2r2)22r2

(2n2r

) (c)

(n2)(n2

2r4)22r4

(2n2r

)

8. Two dice are thrown. Let A be the event that the sum of the faces is odd, B bethe event of at least one ace. Describe the events AB, A B , ABc. Find theirprobabilities assuming that all 36 sample points have equal probabilities.

8/9/2019 PS1Answer

2/2

Answer: A= {12, 21, 41, 14, 23, 32, 16, 61, 25, 52, 34, 43, 36, 63, 45, 54, 56, 65}

B ={11, 12, 21, 13, 31, 14, 41, 15, 51, 16, 61}. P(AB) = |AB|36

= 636

= 16

.P(A B) = P(A) + P(B) P(AB) = 18

36+ 11

36 6

36= 23

36.

ABc=sum of faces is odd and no unit face. P(ABc) = 1236

9. Show that P(ni=1Ai)

ni=1 P(Ai) (n 1)Answer: P(ni=1Ai) = 1 P(

ni=1A

ci) 1

ni=1 P(A

ci) = 1

ni=1(1 P(Ai))

10. An urn contains 3 red balls and 5 blue balls. Calculate the probability that in twodraws the second draw will be a blue ball given the first draw is red.

Answer: 57

.

11. In answering a question on a multiple choice test a student either knows the answeror she guesses. Let p be the probability that she knows the answer and 1 p theprobability that she guesses. Assume that a student who guesses at the answer will

be correct with probability 1m , wherem is the number of multiple choice alternatives.

What is the conditional probability that a student knew the answer to a questiongiven that she answered it correctly?

Answer: This requires Bayes theorem; we will learn Bayes theorem in the comingweek. However, here is the answer:P(answer is correct)=P(Answer is correct |She knows the answer) P(She knows heanswer) +P(Answer is correct |She guesses) P(She guesses) = p + 1p

m

Thus, P(She knows the answer | Answer is correct) = pp+ 1p

m

.

12. IfP(Ac) = 0.3, P(B) = 0.4 and P(A Bc) = 0.5, find P(B|A Bc).

Answer: P(B|A Bc) = P(AB)P(ABc) . P(AB) = P(A) P(A Bc) = 0.7 0.5 = 0.2.P(A Bc) =P( Ac B) =P( (B AB)) = 1 (0.4 0.2) = 0.8.Thus required probability is 1

4.