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.
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.
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).
Kigele yazidi said:
2 years ago
Total number of balls = 8 + 7 + 6,
=21.
Probability not red or green= number of events/sample space.
So, 7รท21 = 1/3.
=21.
Probability not red or green= number of events/sample space.
So, 7รท21 = 1/3.
(3)
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.
Tushar dhok said:
8 years ago
Total no.of ball 8, 7, 6 = 21.
E= event that the ball is neither red nor green.
= event that the ball is red.
n(E) = 8.
P(E) = 8/21.
E= event that the ball is neither red nor green.
= event that the ball is red.
n(E) = 8.
P(E) = 8/21.
(1)
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.
Aruna said:
1 decade ago
Neither red nor blue.
Then why we are take 7 blue balls only.
Can't understand.
Please clear my doubt.
Then why we are take 7 blue balls only.
Can't understand.
Please clear my doubt.
Anitha said:
6 years ago
Total no.of balls =8+7+6=21,
Neither red nor green,
I.e.,we need to take 8,
Therefore p(e)=8/21.
Neither red nor green,
I.e.,we need to take 8,
Therefore p(e)=8/21.
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
Boopathy said:
7 years ago
Probability of selection Red or Green= 8+6=14 / 21 = 2/3.
Then how 1/3?
Then how 1/3?
(1)
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