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.
Angamuthu SURESH said:
1 decade ago
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 red.
Therefore, n(E) = 8.
P(E) = 8/21.
= 21.
Let E = event that the ball drawn is neither red nor green
=event that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.
Farah said:
1 decade ago
Please all students read question carefully what he ask in question. He said that 8 red, 6 green, 7 blue total=21 probability ?
He said neither from red and green so only one option is 7 blue.
7c1 which is n(a)=7 and n(S)=21c1=21
Formula is prob. p(A)=n(A)/n(S)
7/21 = 1/3.
He said neither from red and green so only one option is 7 blue.
7c1 which is n(a)=7 and n(S)=21c1=21
Formula is prob. p(A)=n(A)/n(S)
7/21 = 1/3.
Muhammad Imran Tariq said:
1 decade ago
Sample= 8+7+6=21.
Probability of selection Red and Green= 8+6=14 / 21 = 2/3.
Not selection of Red and Green. Simple subtracts 1 from 2/3.
1 - 2/3= 1/3 (Answer).
Probability of selection Red and Green= 8+6=14 / 21 = 2/3.
Not selection of Red and Green. Simple subtracts 1 from 2/3.
1 - 2/3= 1/3 (Answer).
Mangesh Modiraj said:
1 decade ago
@Nandini & @Monica.
In example clearly mentioned that the ball ins neither red nor green.
i.e blue.
We have 7 blue balls.
Therefore n(E) = 7.
In example clearly mentioned that the ball ins neither red nor green.
i.e blue.
We have 7 blue balls.
Therefore n(E) = 7.
Bibekpangeni said:
1 decade ago
What is the meaning of probability?
Durga said:
1 decade ago
Total balls are - 21
So the probability of getting one ball is -21c1 = 21.
That one ball is neither red nor green means blue ball.
Posing one blue ball from 7 blue balls is 7c1 = 7.
Our required ans=7/21=1/3.
So the probability of getting one ball is -21c1 = 21.
That one ball is neither red nor green means blue ball.
Posing one blue ball from 7 blue balls is 7c1 = 7.
Our required ans=7/21=1/3.
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