Question
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?
Answer: Option B
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
= $${}^{15}\mathop C\nolimits_2 $$
=$$\frac{{15!}}{{2!(15 - 2)!}}$$
=$$\frac{{15!}}{{2! × 13!}}$$
=$$\frac{{15 × 14}}{{1 × 2}}$$
=105
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.
$$ = {}^{10}\mathop C\nolimits_2 $$
= $$\frac{{10!}}{{2!(10 - 2)!}}$$
= $$\frac{{10!}}{{2! × 8!}}$$
= $$\frac{{10 × 9}}{{1 × 2}}$$
= 45
∴ Required probability = $$\frac{{45}}{{105}}$$ = $$\frac{{3}}{{7}}$$
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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
= $${}^{15}\mathop C\nolimits_2 $$
=$$\frac{{15!}}{{2!(15 - 2)!}}$$
=$$\frac{{15!}}{{2! × 13!}}$$
=$$\frac{{15 × 14}}{{1 × 2}}$$
=105
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.
$$ = {}^{10}\mathop C\nolimits_2 $$
= $$\frac{{10!}}{{2!(10 - 2)!}}$$
= $$\frac{{10!}}{{2! × 8!}}$$
= $$\frac{{10 × 9}}{{1 × 2}}$$
= 45
∴ Required probability = $$\frac{{45}}{{105}}$$ = $$\frac{{3}}{{7}}$$
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