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

A. $\frac{4}{15}$

B. $\frac{5}{27}$

C. $\frac{1}{3}$

D. $\frac{2}{445}$

Solution:

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 4 balls such that 2 are red and 2 are green.
Then, n(E) =

(^{2} C_{2} \times^{4} C_{2})

= (1 \times \frac{4 \times 3}{2 \times 1})

= 6
And, n(S) =

^{15} C_{4} =

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

= 1365

∴ P (E) = \frac{n (E)}{n (S)} = \frac{6}{1365} = \frac{2}{455}

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