Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 2)
2.
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Answer: Option
Explanation:
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
| Then, n(S) | = Number of ways of drawing 2 balls out of 7 | |||
| = 7C2 ` | ||||
|
||||
| = 21. |
Let E = Event of drawing 2 balls, none of which is blue.
 n(E) |
= Number of ways of drawing 2 balls out of (2 + 3) balls. | |||
| = 5C2 | ||||
|
||||
| = 10. |
| P(E) = |
n(E) | = | 10 | . |
| n(S) | 21 |
Discussion:
115 comments Page 2 of 12.
Tejeshwari said:
3 years ago
5/7 is it right?
Anyone can explain it, please.
Anyone can explain it, please.
(9)
Lyi said:
4 years ago
The probability of the first ball not being blue is 5/7 and the probability of the second ball also not being blue is 4/6 and this is so because we have already drawn one ball and now the total number of balls got reduced to 6. Now where the 4 comes from? Its because we're assuming that the first ball we've drawn is a non-blue and therefore we have one less ball than 5 to draw from now.
I hope it was helpful.
I hope it was helpful.
(26)
SRZ said:
5 years ago
P1st (not blue)= 1-P(blue) = 1-2/7 = 5/7.
P2nd(not blue)= 1-P(blue) = 1-2/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.
P2nd(not blue)= 1-P(blue) = 1-2/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.
(45)
Thejo said:
5 years ago
We should pick 2 from 7 none of the ball drawn is blue, then the picked balls colours should be red and green. How come it is 10/21?
Here it could be 2C1*3C1 ÷ 7C2 = 6/21 right! Can any1 tell me why it is wrong?
Here it could be 2C1*3C1 ÷ 7C2 = 6/21 right! Can any1 tell me why it is wrong?
(2)
Kanishk said:
5 years ago
@Biswajit.
Firstly the probability of drawing two blue balls will be 2/7*1/6.
Secondly, it won't work by subtracting from 1 since two blue balls and no blue balls aren't the only two cases. There are other case balls with different colours, one of the colours being blue.
Hope you get it now!
Firstly the probability of drawing two blue balls will be 2/7*1/6.
Secondly, it won't work by subtracting from 1 since two blue balls and no blue balls aren't the only two cases. There are other case balls with different colours, one of the colours being blue.
Hope you get it now!
(3)
Biswajit said:
5 years ago
Can anyone tell me if I subtract the probability of getting 2 blue balls i.e 2/7 from 1 then why I can't get the answer?
Please explain me.
Please explain me.
(2)
Sai said:
5 years ago
Total number of balls gives 2+3+2=7 s be the sample space.
number of drawing balls n(s)=2 balls out of 7.
7 c 2 = 7 * 6 = 42/2 = 21.
e = drawing balls none of which is blue,
2 balls out of 2+3=5.
5 c2.
5*4 = 20/2 = 10.
Therefore p n(e)= 10/21.
number of drawing balls n(s)=2 balls out of 7.
7 c 2 = 7 * 6 = 42/2 = 21.
e = drawing balls none of which is blue,
2 balls out of 2+3=5.
5 c2.
5*4 = 20/2 = 10.
Therefore p n(e)= 10/21.
(5)
Gufran said:
5 years ago
(2R+3G)c2/7c2.
5c2/7c2.
5 * 2/2/7 * 6/2.
(20/2)/(42/2).
10/21 is the final answer
5c2/7c2.
5 * 2/2/7 * 6/2.
(20/2)/(42/2).
10/21 is the final answer
(1)
Naseem Ullah Khan said:
5 years ago
Simple explanation, well done, thanks @Krish.
(2)
Krish said:
5 years ago
Using conceptual method rather than going by formula gives a better feel for what is happening.
P1 = probability of 1st ball not blue = 5/7 as 5 are non-blue.
P2 = probability of 2nd ball not blue = 4/6 as 1 non-blue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.
P1 = probability of 1st ball not blue = 5/7 as 5 are non-blue.
P2 = probability of 2nd ball not blue = 4/6 as 1 non-blue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.
(10)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers