Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 3)
3.
In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
Answer: Option
Explanation:
Total number of balls = (8 + 7 + 6) = 21.
| Let E | = event that the ball drawn is neither red nor green |
| = event that the ball drawn is blue. |
n(E) = 7.
| P(E) = |
n(E) | = | 7 | = | 1 | . |
| n(S) | 21 | 3 |
Discussion:
36 comments Page 4 of 4.
S.Sakthivel said:
1 decade ago
Because neither which is indicates that. So that the possibility for the red and green is not considered. Thus simultaneously the blue ball is taken under consideration.
Nandini said:
1 decade ago
How n(E) =7?
Sonu said:
1 decade ago
If at least one is blue means what is the answer?
Vasu said:
1 decade ago
How the n(e) come as 7?
Rajeev R T said:
1 decade ago
Event that the ball is neither red nor green, then you are left out with 7 blue balls, and out 7 blue balls you need one ball.
Then it is 7C1 ways... i.e = 7 ways.
Then it is 7C1 ways... i.e = 7 ways.
Monika said:
1 decade ago
HOW n(E)=7?
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