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1. A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue?

A. $\frac{137}{143}$

B. $\frac{18}{455}$

C. $\frac{9}{91}$

D. $\frac{2}{5}$

Solution:

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 5 balls out of 9 non-blue balls.

=^{9} C_{5} =^{9} C_{(9 - 5)} =^{9} C_{4}

= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}

= 126
And,
n(S) =

^{15} C_{5} =

\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}

= 3003

∴ P (E) = \frac{n (E)}{n (S)} =

\frac{126}{3003} = \frac{6}{143}

∴ Required Probability

\begin{aligned} = (1 - \frac{6}{143}) \\ = \frac{137}{143} \end{aligned}

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