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  
= ^{7}C_{2} `  


= 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.  
= ^{5}C_{2}  


= 10. 
P(E) =  n(E)  =  10  . 
n(S)  21 
Discussion:
106 comments Page 1 of 11.
SRZ said:
3 years ago
P1st (not blue)= 1P(blue) = 12/7 = 5/7.
P2nd(not blue)= 1P(blue) = 12/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.
P2nd(not blue)= 1P(blue) = 12/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.
(12)
Lyi said:
3 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 nonblue and therefore we have one less ball than 5 to draw from now.
I hope it was helpful.
I hope it was helpful.
(8)
Krish said:
3 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 nonblue.
P2 = probability of 2nd ball not blue = 4/6 as 1 nonblue 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 nonblue.
P2 = probability of 2nd ball not blue = 4/6 as 1 nonblue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.
(6)
Tejeshwari said:
11 months ago
5/7 is it right?
Anyone can explain it, please.
Anyone can explain it, please.
(3)
Ronak Mistry said:
3 months ago
Total Balls = 7.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.
(3)
Biswajit said:
3 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)
Kanishk said:
3 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!
(2)
Thejo said:
3 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 Ã· 7C2 = 6/21 right! Can any1 tell me why it is wrong?
Here it could be 2C1*3C1 Ã· 7C2 = 6/21 right! Can any1 tell me why it is wrong?
(2)
Harry said:
5 years ago
Can someone help me with this part of the question:
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7.
= 7C2 `
= (7 x 6).
(2 x 1),
= 21.
I don't understand what 7C2 means or how we get 7x6 or 2x1 then divided.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7.
= 7C2 `
= (7 x 6).
(2 x 1),
= 21.
I don't understand what 7C2 means or how we get 7x6 or 2x1 then divided.
(1)
Naseem Ullah Khan said:
3 years ago
Simple explanation, well done, thanks @Krish.
(1)
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