Question

There are two coins, one is fair and one is biased. The biased coin has a probability of landing on heads equal to 4/5. One of the coins is chosen at random (50-50), and is flipped repeatedly until it lands on tail. If it landed on heads 4 times before landing on tails, what is the posterior probability that coin chosen was biased?

Answer 1

Answer:

72.4%

Step-by-step explanation:

The probability of A occurring given that B occurs = the probability of both A and B / the probability of B

P(A|B) = P(A∩B) / P(B)

This can be rearranged as:

P(A∩B) = P(B) P(A|B)

In this case:

A = biased coin is chosen

~A = fair coin is chosen

B = 4 heads then 1 tail

First, let's find P(A∩B).

P(A∩B) = P(B) P(A|B)

P(A∩B) = ½ × ₅C₄ (⅘)⁴ (⅕)¹

P(A∩B) = 0.2048

Next, find P(~A∩B).

P(~A∩B) = P(B) P(~A|B)

P(~A∩B) = ½ × ₅C₄ (½)⁴ (½)¹

P(~A∩B) = 0.078125

Therefore, the probability that the coin is biased is:

P = P(A∩B) / (P(A∩B) + P(~A∩B))

P = 0.2048 / (0.2048 + 0.078125)

P = 0.723866749

The probability is approximately 72.4%.