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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 in colour?
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 in colour?
1). $\frac{1}{7}$
2). $\frac{3}{7}$
3). $\frac{2}{7}$
4). $\frac{5}{14}$
1). $\frac{1}{7}$
2). $\frac{3}{7}$
3). $\frac{2}{7}$
4). $\frac{5}{14}$
This Question has 2 answers.
A bag contains 4 red, 5 yellow, and 6 pink balls.
Total number of balls = $4 + 5 + 6 = 15$
Total ways to choose 2 balls from 15:
$ \binom{15}{2} = \frac{15!}{2!(15-2)!} = \frac{15 \times 14}{2} = 105 $
Number of non-yellow balls = red balls + pink balls = $4 + 6 = 10$
Ways to choose 2 balls from these 10:
$ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2} = 45 $
Probability that none of the balls drawn are yellow:
$ P(\text{none is yellow}) = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{45}{105} $
Simplifying:
$ P(\text{none is yellow}) = \frac{3}{7} $
Final Answer:
$ \boxed{\frac{3}{7}} $
Total number of balls = $4 + 5 + 6 = 15$
Total ways to choose 2 balls from 15:
$ \binom{15}{2} = \frac{15!}{2!(15-2)!} = \frac{15 \times 14}{2} = 105 $
Number of non-yellow balls = red balls + pink balls = $4 + 6 = 10$
Ways to choose 2 balls from these 10:
$ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2} = 45 $
Probability that none of the balls drawn are yellow:
$ P(\text{none is yellow}) = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{45}{105} $
Simplifying:
$ P(\text{none is yellow}) = \frac{3}{7} $
Final Answer:
$ \boxed{\frac{3}{7}} $
Solution
Number of non-yellow balls = 4 + 6 = 10
Required Probability = $\frac{^{10}C_2}{^{15}C_2} = \frac{3}{7}$
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