Urn has 5 white and 7 black balls. Urn has 3 white and 12 black balls. We flip a fair coin. If the outcome is heads, then a ball from urn is selected, whereas if the outcome is tails, then a ball from urn is selected. Suppose that a white ball is selected. What is the probability that the coin landed tails?

Urn A has 5 white and 7 black balls. Urn B has 3 white and 12 black balls. We flip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Suppose that a white ball is selected. What is the probability that the coin landed tails?

Solution

Let the probability of that a head turns up and urn A is selcted be P(A) = (1/2)

Probability that a tail turns up and urn B is selected = P(B) = (1/2)

Probability that a white ball is picked = P(W)

The probability that a white ball is picked given that the coin toss gives a head and urn A is selected = P(W|A) = (5/12)

The probability that a white ball is picked given that the coin toss gives a tail and urn B is selected = P(W|B) = (3/15) = (1/5)

We now require the probability that the coin lands on a tail and urn B is selected given that a white ball is picked, P(B|W)

Note that the conditional probability P(X|Y) is expressed mathematically as

P(X|Y) = P(X n Y) ÷ P(Y)

And P(X n Y) = P(X|Y) × P(Y)

Hence, the required probability

P(B|W) = P(B n W) ÷ P(W)

Although, we do not have the probabilities P(B n W) and P(W), we can calculate them

P(B n W) = P(W n B) = P(W|B) × P(B) = (1/5) × (1/2) = (1/10)

P(W) = P(W n A) + P(W n B) (Since the events A and B are mutually exclusive)

P(W n A) = P(W|A) × P(A) = (5/12) × (1/2) = (5/24)

P(W n B) = (1/10)

P(W) = (5/24) + (1/10) = (37/120)

P(B|W) = P(B n W) ÷ P(W) = (1/10) ÷ (37/120) = (12/37) = 0.3243

Hope this Helps!!!