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Probability

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Question : 72 of 127

Marks: +1, -0

There are three boxes with the following composition:

Box 1: 5 Red +7 White +6 Blue balls

Box II: 4 Red +8 White +6 Blue balls

Box III: 3 Red +4 White +2 Blue balls

If one ball is drawn (from each) at random, then what is the probability that they would be of same colour?

$\frac{89}{729}$
$\frac{97}{729}$
$\frac{82}{729}$
$\frac{23}{32}$

Solution:

Concept:

Probability of independent events; sum probabilities for each colour.

Explanation:

Total balls: Box I = 18, Box II = 18, Box III = 9.

All red:

\frac{5}{18} \times \frac{4}{18} \times \frac{3}{9} = \frac{60}{2916} = \frac{15}{729}

.

All white:

\frac{7}{18} \times \frac{8}{18} \times \frac{4}{9} = \frac{224}{2916} = \frac{56}{729}

.

All blue:

\frac{6}{18} \times \frac{6}{18} \times \frac{2}{9} = \frac{72}{2916} = \frac{18}{729}

.

Total probability =

\frac{15}{729} + \frac{56}{729} + \frac{18}{729} = \frac{89}{729}

.

Answer:

\frac{89}{729}

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