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 3 of 4.
Ravindra said:
6 years ago
How can p(e) =7?
Because neither red nor green clearly means its blue.
Because neither red nor green clearly means its blue.
(1)
Ramachandran Nichite said:
9 years ago
Hi @Obiano.
Total ball = 5 + 3 + 2 = 10.
Choose yellow ball = 5/10.
Total ball = 5 + 3 + 2 = 10.
Choose yellow ball = 5/10.
Faffy said:
9 years ago
Does this mean that we are excluding green and red?
Sonu said:
1 decade ago
If at least one is blue means what is the answer?
Hina said:
1 decade ago
What is the meaning that die is fair or not?
Otuoboizy said:
10 years ago
Total numbers of balls is 21 and cl is 7.
Sravani said:
1 decade ago
n(E) =7 because there are 7 blue balls.
Bibekpangeni said:
1 decade ago
What is the meaning of probability?
SRINU said:
1 decade ago
Total balls-21.
7C1 = 7.
7/21.
7C1 = 7.
7/21.
Jaiprakash said:
9 years ago
@Obiano.
The answer is 1/2.
The answer is 1/2.
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