A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Options:
| A. | \(\frac{10}{21}\) |
| B. | \(\frac{11}{21}\) |
| C. | \(\frac{2}{7}\) |
| D. | \(\frac{5}{7}\) |
Answer: Option A
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2
= \(\frac{(7\times6)}{(2\times1)}\)
=21.
Let E = Event of drawing 2 balls, none of which is blue.
So, n(E) = = Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
= \(\frac{(5\times4)}{(2\times1)}\)
=10.
So, \(P(E) = \frac{n(E)}{n(S)}=\frac{10}{21}\)
1 Answers
When i ask new question how to ask?
Submit Solution