Question

A box $B_1$ contains $1$ white ball, $3$ red balls and $2$ black balls. Another box $B_2$ contains $2$ white balls, $3$ red balls and $4$ black balls. A third box $B_3$ contains $3$ white balls, $4$ red balls and $5$ black balls.

If 1 ball is drawn from each of the boxes $B_1,B_2$ and $B_3$, the probability that all 3 drawn balls are of the same colour is

• A. $\frac{82}{648}$
• B. $\frac{90}{648}$
• C. $\frac{558}{648}$
• D. $\frac{566}{648}$

Answer 1

The correct option is A $\frac{82}{648}$

$P\left(required\right)$=$P\left(allarewhite\right)+P\left(allarered\right)+P\left(allareblack\right)$

$=\frac{1}{6}\times \frac{2}{9}\times \frac{3}{12}+\frac{3}{6}\times \frac{3}{9}\times \frac{4}{12}+\frac{2}{6}\times \frac{4}{9}\times \frac{5}{12}$

$=\frac{6}{648}+\frac{36}{648}+\frac{40}{648}=\frac{82}{648}$