Question

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?

Short answer

The probability that the coin landed tails is $\frac{12}{37}$.

Step-by-step solution

Given information

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.

Solution

Urn $A$$5$ white

$7Black\Rightarrow P\left(white\right)=\frac{5}{12}$

Urn $B$$3$ White

$12Black\Rightarrow P\left(white\right)=\frac{3}{15}$

$P\left(\text{Tail | white}\right)=\frac{\frac{1}{2}\times \frac{3}{15}}{\frac{1}{2}\times \frac{3}{15}+\frac{1}{2}\times \frac{5}{12}}$

$=\frac{3}{30}\times \frac{240}{74}$

$=\frac{12}{37}$

Final answer

The probability that the coin landed tails is$\frac{12}{37}$.