Question

Suppose that we have two urns, 1 and 2, each with two drawers. Urn 1 has a gold coin in one drawer and a silver coin in the other drawer which urn 2 has a gold coin is each drawer. One urn is chosen at random; then a drawer is chosen at random from the chosen urn. The coin found in drawer turns out to be gold . What is the probability that the coin came from urn 2?

Answer 1

Answer on Question 37427 – Math – Statistics and Probability

We will need the following formula

$$P \left(B _ {i} \mid A\right) = \frac {P \left(B _ {i}\right) P \left(A \mid B _ {i}\right)}{\sum_ {k = 1} ^ {n} P \left(B _ {k}\right) P \left(A \mid B _ {k}\right)}$$

This formula is a special case of the well-known Bayes' Theorem.

Let the events be as followed:

$B_{1}$ : We choose the first urn;

$B_{2}$ : We choose the second urn;

$A$ : We found the gold coin.

We are searching for $P(B_{2}|A)$.

In our problem,

$$P \left(B _ {1}\right) = P \left(B _ {2}\right) = \frac {1}{2},$$ $$P (A | B _ {1}) = \frac {1}{2},$$ $$P (A | B _ {2}) = 1$$

Using $(^{*})$, we get

$$P (B _ {2} | A) = \frac {P (B _ {2}) P (A | B _ {2})}{P (B _ {1}) P (A | B _ {1}) + P (B _ {2}) P (A | B _ {2})} = \frac {\frac {1}{2} * 1}{\frac {1}{2} * \frac {1}{2} + \frac {1}{2} * 1} = \frac {2}{3}.$$

Answer: 2/3.