Question

Suppose there are two coins. One is a standard fair coin, so that P(heads)=0.50. The other one is a two-sided coin, so that P(heads)=1. You draw one of the two coins at random and toss it. It results in heads. Given that observation... (a) Compute the probability that you have selected a fair coin. (1pt) (b) What is the probability that the next toss will result in heads too? (1pt) (c) If the next toss results in heads as well, what is the updated probability that you have selected a fair coin? (1pt)

Answer 1

a) As we are given here that we are selecting the two coins randomly, therefore:
P(Coin1) = P(Coin2) = 0.5

Also, we are given here that:
P(H | Coin1) = 0.5 and P(H | Coin2) = 1

Using law of total probability, we have here:
P(H) = P(H | Coin1)P(Coin1) + P(H | Coin2) P(Coin2) = 0.5*(0.5 + 1) = 0.75

a) Given that a heads is the outcome, the probability that we selected a fair coin is computed using Bayes theorem as:
P(Coin1 | H) = P(H | Coin1)P(Coin1) / P(H) = 0.5*0.5 / 0.75 = 1/3

Therefore 1/3 is the required probability here.

b) Probability that the next toss will result in heads too is computed here as:
= 0.5*P(Coin1 | H) + 1*P(Coin2 | H)

= 0.5*(1/3) + 1*(2/3)

= 2.5/3 = 5/6

Therefore 5/6 is the required probability here.

c) Given that the netx toss is also heads, probability that the selected coin is fair is computed here as:

P(HH | Coin1) = 0.5² = 0.25,
P(HH | Coin2) = 1

Therefore,
P(HH) = 0.25*0.5 + 1*0.5 = 0.625

Therefore P(Coin1 | HH) = 0.25*0.5 / 0.625 = 0.2

Therefore 0.2 is the required probability here.