A bag contains 3 red, 4 white and 5 blue balls (All balls are different). If two balls are drawn at random, then the probability that they are of different colours is:

\frac{47}{66}

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\frac{10}{33}

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\frac{5}{22}

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N o n e

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Solution

The correct option is A $\frac{47}{66}$
We have the following pattern:
$\begin{matrix} I & r e d, & w h i t e & P_{I} = 3 \times \frac{4}{^{12} C_{2}} I I & r e d, & b l u e & P_{I I} = 3 \times \frac{5}{^{12} C_{2}} I I I & b l u e, & w h i t e & P_{I I I} = 4 \times \frac{5}{^{12} C_{2}} \end{matrix}$
Since all these cases are exclusive, so the required probability = $\frac{(12 + 15 + 20)}{^{12} C_{2}} = \frac{47}{66}$

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