Question

In a bag, there are 6 balls of which 3 are white and 3 are black. 6 balls are drawn successively $\left(i\right)$ without replacement, $\left(ii\right)$ with replacement. What is the chance that the colors are alternate?

• A. $\frac{1}{20}$; $\frac{1}{64}$
• B. $\frac{1}{10}$; $\frac{1}{64}$
• C. $\frac{1}{20}$; $\frac{1}{32}$
• D. $\frac{1}{10}$; $\frac{1}{32}$

Answer 1

The correct option is D $\frac{1}{10}$; $\frac{1}{32}$

$\left(i\right)$ Without replacement
$P\left\{\text{six balls drawn follow the pattern}WBWBWB\right\}$
$=\frac{3}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{2}{3}\times \frac{1}{2}\times \frac{1}{1}=\frac{1}{20}$
$P\left\{\text{six balls drawn follow the pattern}BWBWBW\right\}$
$=\frac{3}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{2}{3}\times \frac{1}{2}\times \frac{1}{1}=\frac{1}{20}$
Hence, the probability that the colors of the balls are alternate is the probability that either the pattern $WBWBWB$ or $BWBWBW$ is obtained, which is given by $\frac{1}{20}+\frac{1}{20}=\frac{1}{10}$ (this is without replacement)

$\left(ii\right)$ With replacement
Here, the probability of getting the pattern $WBWBWB$ is ${\left(\frac{3}{6}\right)}^6={\left(\frac{1}{2}\right)}^6=\frac{1}{64}$
$\therefore$ $P\left\{\text{colors of the balls are alternate}\right\}=\frac{1}{64}+\frac{1}{64}$
$=\frac{1}{32}$