Question

3 Suppose that a box contains five coins, and that for each coin there is a different probability that a head will be obtained when the coin is tossed. Let pi denote the probability of a head when the ith coin is tossed, where i 1,2,3, 4,5]. Suppose that a (8 marks) Suppose that one coin is selected at random from the the probability that the ith coin was selected? Note that i b (8 marks) If the same coin were tossed again, what would be the box and when it is tossed once, a head is obtained. What is probability of obtaining another head? Hint: We define the following events C - (the ith coin was selected, i -1,2, 3,4,5), H-head on the jth toss, j-1,2) T tail on the jth toss, j-1,2]. In your calculations, you may use the identities written below after replacing the symbol "?" with the correct quantities P(H2| Hi) P( H2IC" n H)P(CIHİ ) [Partition Th. with extra conditioning] P(CİHı) formula cond. probl P(C H) formula cond. prob.] PH2C)PHC)PH) (conditional independencel c (8 marks) If a tail had been obtained on the first toss of the selected coin and the same coin were tossed again, what would be the probability of obtaining a head on the second toss?

Answer 1

SOLUTION

Let

P(i) = the probability that the ith coin is selected
P(head) =a head results from the toss

Here,

P(i|head) = P(i and head) / P(head)

Here,

P(head) = P(1) P(head|1) + P(2) P(head|2) + P(3) P(head|3) + P(4) P(head|4) + P(5) P(head|5)

As P(i) = 1/5,

P(head) = (1/5)(0) + (1/5)(1/4) +(1/5)(1/2) + (1/5)(3/4) + (1/5)(1) = 1/2

Also,

P(i and head) = P(i) P(head|i) = (1/5) p_i

Thus,

P(i|head) = (1/5) p_i / (1/2)

P(i|head) = (2/5) p_i [ANSWER]

*************************

B)

If the same coin is tossed, then we already know what coin it is, and its probability.

Hence,

P(head|i) = p_i [ANSWER]