Two balls are selected at random one by one without replacement from a bag containing \(4\) white and \(6\) black balls.…

The probability that the first ball is black is \(\frac{6}{10}\)

After selecting a black ball, there are \(5\) black balls

left out of \(9\) total balls.

So, the probability that the second ball is black is \(\frac{5}{9}\)

\(P(A\cap B)=\frac{6}{10}\cdot \frac{5}{9}=\frac{30}{90}=\frac{1}{3}\)

Now, \(P(B)=\frac{6}{10}\times \frac{5}{9}+\frac{4}{10}\times \frac{6}{9}=\frac{1}{3}+\frac{4}{15}=\frac{5}{15}+\frac{4}{15}\)\(=\frac{9}{15}=\frac{3}{5}\)

By conditional probability,

\(P\left(A∣B\right)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3}\cdot \frac{5}{3}\)\(=\frac{5}{9}\)

The probability \(\text{ }P(A∣B)=\frac{5}{9}\text{.}\)

\(\text{Here, }m=5\text{ and }n=9\text{, and }\gcd (5,9)=1\text{. }\)

\(m+n=5+9=14\)