A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls. If 1 ball is drawn from each of the boxes B1, B2 and B3 the probability that all 3 drawn balls are of the same color is

$\frac{82}{648}$

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\frac{90}{648}

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\frac{558}{648}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{566}{648}$

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Solution

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

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Similar questions

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 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 color is