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1. 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?
- A. 5/7
- B. 5/7
- C. 5/7
- D. 5/7
Answer: Option C
Explanation:
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
n(S) = 7C2
n(S) = (7 × 6)/(2 × 1)
n(S) = 21
Let E = Event of 2 balls, none of which is blue
∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
n(E) = 5C2
n(E) = (5 × 4)/(2 × 1)
n(E) = 10
∴P(E) = n(E)/n(S) = 10/21
= (2 + 3 + 2)
= 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
n(S) = 7C2
n(S) = (7 × 6)/(2 × 1)
n(S) = 21
Let E = Event of 2 balls, none of which is blue
∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
n(E) = 5C2
n(E) = (5 × 4)/(2 × 1)
n(E) = 10
∴P(E) = n(E)/n(S) = 10/21
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