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1. 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?

A. $\frac{3}{455}$

B. $\frac{1}{5}$

C. $\frac{4}{5}$

D. $\frac{44}{91}$

Solution:

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 3 non-yellow balls
Then, n(E) =

^{12} C_{3}

= \frac{12 \times 11 \times 10}{3 \times 2 \times 1}

= 220
Also, n(S) =

^{15} C_{3}

= \frac{15 \times 14 \times 13}{3 \times 2 \times 1}

= 455

∴ P (E) = \frac{n (E)}{n (S)} = \frac{220}{455} = \frac{44}{91}

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