Question

A box contains $5$ black, $4$ white and $6$ red balls. Two balls are drawn one after another without replacement. The probability that the first will be white and the second will be black is

• A. $\frac{4}{21}$
• B. $\frac{1}{21}$
• C. $\frac{2}{21}$
• D. $\frac{5}{12}$

Answer 1

The correct option is C $\frac{2}{21}$

Let $A=$ first will be white ball, and
$B=$ second will be black
Required probability:
$P(A\cap B)=P\left(A\right)\cdot P\left(\frac{B}{A}\right)=\frac{4}{15}\times \frac{5}{14}=\frac{2}{21}$