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?
 1 3
 3 4
 7 19
 8 21
 9 21
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 1 of 4.

Monika said:   1 decade ago
HOW n(E)=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.

Vasu said:   1 decade ago
How the n(e) come as 7?

Sonu said:   1 decade ago
If at least one is blue means what is the answer?

Nandini said:   1 decade ago
How n(E) =7?

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.

Sravani said:   1 decade ago
n(E) =7 because there are 7 blue balls.

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.

Sahi said:   1 decade ago
Hi aruna I think you didn't read the question properly it is clearly mentioned that neither red nor green, so the remaining are blue. Hence we hve to consider blue balls. Hope you got it :).

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

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