A bag contains 3 red, 6 white and 7 black. Two balls are drawn at random. Probability that both are black?

\frac{1}{8}

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\frac{7}{40}

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\frac{12}{40}

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\frac{13}{40}

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Solution

The correct option is B $\frac{7}{40}$
$H e r e 3 r e d, 6 w h i t e a n d 7 b l a c k b a l l s$

$P = \frac{^{7} C_{2}}{^{16} C_{2}} = \frac{21}{120} = \frac{7}{40}$

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