# Aptitude - Probability - Discussion

@ : Home > Aptitude > Probability > General Questions - Discussion

"Do not wait for leaders; do it alone, person to person."
- Mother Teresa
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?

[A].
 1 3
[B].
 3 4
[C].
 7 19
[D].
 8 21
[E].
 9 21

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

 Monika said: (Fri, Mar 11, 2011 08:44:01 AM) HOW n(E)=7?

 Rajeev R T said: (Mon, Apr 25, 2011 11:09:16 AM) 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: (Wed, Jul 20, 2011 12:23:39 PM) How the n(e) come as 7?

 Sonu said: (Sat, Aug 6, 2011 02:27:35 PM) If at least one is blue means what is the answer?

 Nandini said: (Mon, Aug 22, 2011 08:24:46 AM) How n(E) =7?

 S.Sakthivel said: (Tue, Aug 30, 2011 12:18:51 AM) 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: (Sat, Nov 12, 2011 10:18:38 PM) n(E) =7 because there are 7 blue balls.

 Aruna said: (Thu, Dec 22, 2011 02:22:42 PM) Neither red nor blue. Then why we are take 7 blue balls only. Can't understand. Please clear my doubt.

 Sahi said: (Fri, Dec 30, 2011 12:29:50 AM) 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: (Sat, Feb 25, 2012 11:53:59 AM) Hear there are 1 randomly selected means green 8 blue 6 average of 2colors so 8+6=14 14/2=7

 Sree said: (Tue, Mar 6, 2012 09:45:30 AM) There is a die with 10 faces. It is not known that fair or not. 2 captains want to toss die for batting selection. What is the possible solution among the following? a) If no. is odd it is head, if no. is even it is tail b) If no. is odd it is tail, if no. is even it is head c) Toss a die until all the 10 digits appear on top face. And if first no. in the sequence is odd then consider it as tail. If it is even consider it as head. what is the exact solution

 Hina said: (Thu, May 10, 2012 11:54:32 AM) What is the meaning that die is fair or not?

 Angamuthu Suresh said: (Wed, Oct 17, 2012 10:34:41 AM) 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.

 Farah said: (Wed, Dec 5, 2012 11:27:50 PM) 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.

 Muhammad Imran Tariq said: (Mon, Dec 24, 2012 01:25:04 AM) 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).

 Mangesh Modiraj said: (Mon, Jul 8, 2013 09:33:05 AM) @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.

 Bibekpangeni said: (Tue, Feb 4, 2014 05:37:48 AM) What is the meaning of probability?

 Durga said: (Sat, Apr 26, 2014 04:18:41 PM) 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.

 Srinu said: (Sat, Feb 7, 2015 03:01:14 PM) Total balls-21. 7C1 = 7. 7/21.

 Otuoboizy said: (Fri, Oct 23, 2015 02:57:52 AM) Total numbers of balls is 21 and cl is 7.

 Knowledgeispower said: (Sat, Feb 6, 2016 09:38:12 AM) Using the same problem, solve for the probability of pulling a RED ball on 2 draws without replacing the ball after each draw.