Question
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?
Answer: Option D
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) = $$\left( {{}^2\mathop C\nolimits_2 \times {}^4\mathop C\nolimits_2 } \right)$$ $$ = \left( {1 \times \frac{{4 \times 3}}{{2 \times 1}}} \right)$$ = 6
And, n(S) = $${}^{15}\mathop C\nolimits_4 = $$ $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$ = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{1365}} = \frac{2}{{455}}$$
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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) = $$\left( {{}^2\mathop C\nolimits_2 \times {}^4\mathop C\nolimits_2 } \right)$$ $$ = \left( {1 \times \frac{{4 \times 3}}{{2 \times 1}}} \right)$$ = 6
And, n(S) = $${}^{15}\mathop C\nolimits_4 = $$ $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$ = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{1365}} = \frac{2}{{455}}$$
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