Question

A box $B_1$ contains $1$white balls, $3$ 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 color, is

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

Answer 1

The correct option is A

$\frac{82}{648}$

The explanation for the correct option:

Step 1: Express the given data:

In Box $B_1$:

Number of a white ball $=1$

Number of red balls $=3$

Number of black balls $=2$

Therefore, total balls in bag $B_1$$=1+3+2=6$

In Box $B_2$:

Number of a white ball $=2$

Number of red balls $=3$

Number of black balls $=4$

Therefore, total balls in bag $B_2$$=2+3+4=9$

In Box $B_3$:

Number of a white ball $=3$

Number of red balls $=4$

Number of black balls $=5$

Therefore, total balls in bag $B_3$$=3+4+5=12$

Step 2: Finding the probability:

The probability that it is white in all the boxes, P (All are white) $=\frac{1}{6}\times \frac{2}{9}\times \frac{3}{12}$

$=\frac{6}{648}$

The probability that it is red in all the boxes, P(All are red) $=\frac{3}{6}\times \frac{3}{9}\times \frac{4}{12}$

$=\frac{36}{648}$
The probability that it is black in all the boxes, P(All are black) $=\frac{2}{6}\times \frac{4}{9}\times \frac{5}{12}$

$=\frac{40}{648}$

Thus the probability that all the three balls are drawn are of the same color is

$$\begin{gathered} =\left(\frac{6}{648}+\frac{36}{648}+\frac{40}{648}\right) \\ =\frac{\mathbf{82}}{\mathbf{648}} \end{gathered}$$

Hence, option (A) is the correct answer.