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1. A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random. What is the probability that none of the balls drawn are yellow?

A. $\frac{1}{7}$

B. $\frac{3}{7}$

C. $\frac{2}{7}$

D. $\frac{5}{14}$

Solution:

Number of red balls = 4
Number of yellow balls = 5
Number of pink balls = 6
Total balls = 4 + 5 + 6 = 15
Total possible outcomes = selection of 2 balls out of 15 balls
=

^{15} C_{2}

\frac{15!}{2! (15 - 2)!}

\frac{15!}{2! \times 13!}

\frac{15 \times 14}{1 \times 2}

=105
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.

=^{10} C_{2}

\frac{10!}{2! (10 - 2)!}

\frac{10!}{2! \times 8!}

\frac{10 \times 9}{1 \times 2}

= 45
∴ Required probability =

\frac{45}{105}

\frac{3}{7}

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