Question

Alice has two coins. The probability of Heads for the first coin is 1/4, and the probability of Heads for the second is 3/4. Other than this difference, the coins are indistinguishable. Alice chooses one of the coins at random and sends it to Bob. The random selection used by Alice to pick the coin to send to Bob is such that the first coin has a probability p of being selected. Assume that 0<p<1. Bob tries to guess which of the two coins he received by tossing it 3 times in a row and observing the outcome. Assume that for any particular coin, all tosses of that coin are independent.

For this part, assume that p=3/4.

a.What is the probability that Bob will guess the coin correctly using the decision rule from part 2?

b.Suppose instead that Bob tries to guess which coin he received without tossing it. He still guesses the coin in order to minimize the probability of error. What is the probability that Bob will guess the coin correctly under this scenario?

Answer 1

a) Bob can be correct when the actual coin is C1 or C2:

P(Bob correct)=P(Bob correct|C1)P(C1)+P(Bob correct|C2)P(C2)

We can calculate one of them as follows (P(C1)=3/4,P(C2)=1/4:

P(Bob correct|C1) = P(k≤2|C1) = 1−P(k=3|C1) = 1−(1/4)^3 = 0.984375

simillarly, P(Bob correct|C2)=0.140625

P(Bob correct)=P(Bob correct|C1)P(C1)+P(Bob correct|C2)P(C2)

= 0.984375 * 0.75 + 0.140625 * 0.25

= 0.84375

b) Because Bob will always decide Coin 1 (since the prior is bigger). And, probability of Coin 1 is 3/4. So, 3/4 of the time Bob will be correct.

P(correct) = P(coin=1∣p=3/4)

=1⋅3/4

=3/4

=0.75