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 2 of 4.
SRINU said:
1 decade ago
Total balls-21.
7C1 = 7.
7/21.
7C1 = 7.
7/21.
Rohit said:
6 years ago
How n(E) =7?
Jaiprakash said:
8 years ago
@Obiano.
The answer is 1/2.
The answer is 1/2.
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.
Obiano kingseley said:
9 years ago
A bag contains 5yellow balls, 3green balls, and 2red balls, a ball is chosen at random from the bag. What is the probability that the chosen ball is yellow?
Can anyone answer this?
Can anyone answer this?
Faffy said:
9 years ago
Does this mean that we are excluding green and red?
Knowledgeispower said:
9 years ago
Using the same problem, solve for the probability of pulling a RED ball on 2 draws without replacing the ball after each draw.
Otuoboizy said:
10 years ago
Total numbers of balls is 21 and cl is 7.
S.Ranga reddy said:
1 decade ago
Hear there are 1 randomly selected means
green 8 blue 6
average of 2colors
so 8+6=14
14/2=7
green 8 blue 6
average of 2colors
so 8+6=14
14/2=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.
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