Question

Each of 2 cabinets identical in appearance has 2 drawers. Cabinet A contains a silver coin in each drawer, and cabinet B contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?

Answer 1

Answer:

The probability that there is a silver coin in the other drawer is $\frac{2}{3}$

Step-by-step explanation:

Given : Each of 2 cabinets identical in appearance has 2 drawers. Cabinet A contains a silver coin in each drawer, and cabinet B contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found.

To find : What is the probability that there is a silver coin in the other drawer?

Solution :

Let A be the event that cabinet A is chosen.

Let B be the event that cabinet B is chosen

Let E be the event that silver coin is chosen.

Each of 2 cabinets identical in appearance has 2 drawers.

So, $P(A)=P(B)=\frac{1}{2}$

Cabinet A contains a silver coin in each drawer,

i.e. $P(E|A)=1$

Cabinet B contains a silver coin in one of its drawers and a gold coin in the other,

i.e. $P(E|B)=\frac{1}{2}$

The probability that there is a silver coin in the other drawer is given by,

Applying Bayes theorem,

$P(A|E)=\frac{P(E|A)P(A)}{P(E|A)P(A)+P(E|B)P(B)}$

$P(A|E)=\frac{(1)(\frac{1}{2})}{(1)(\frac{1}{2})+(\frac{1}{2})(\frac{1}{2})}$

$P(A|E)=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{4}}$

$P(A|E)=\frac{\frac{1}{2}}{\frac{3}{4}}$

$P(A|E)=\frac{2}{3}$

Therefore, the probability that there is a silver coin in the other drawer is $\frac{2}{3}$