Question
A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow?
Answer: Option D
Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 3 non-yellow balls
Then, n(E) = $${}^{12}\mathop C\nolimits_3 $$ $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$ = 220
Also, n(S) = $${}^{15}\mathop C\nolimits_3 $$ $$ = \frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$ = 455
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{220}}{{455}} = \frac{{44}}{{91}}$$
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Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 3 non-yellow balls
Then, n(E) = $${}^{12}\mathop C\nolimits_3 $$ $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$ = 220
Also, n(S) = $${}^{15}\mathop C\nolimits_3 $$ $$ = \frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$ = 455
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{220}}{{455}} = \frac{{44}}{{91}}$$
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