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:
116 comments Page 2 of 12.
Sneha said:
1 year ago
Can anyone explain this in the short method?
(8)
Ahmed Raafat said:
2 years ago
7 CHOOSE 2 can also be denoted as 7C2.
7 is the total number of distinct elements (n),
2 is the number of elements drawn or chosen at a time (k),
21 is the total number of possible combinations (C).
nCk = n! / (k! * (n - k)!)
7C2 =7! / (2! x 5!) = 21.
7 is the total number of distinct elements (n),
2 is the number of elements drawn or chosen at a time (k),
21 is the total number of possible combinations (C).
nCk = n! / (k! * (n - k)!)
7C2 =7! / (2! x 5!) = 21.
(7)
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)
IMx said:
1 month ago
If we did it your way, assuming:
Probability that the first ball is not blue = 5/7,
Probability that second ball is not blue = 4/6 = 2/3,
Then:
P(no blue in both draws) = 5/7 × 4/6 = 20/42 = 10/21.
Probability that the first ball is not blue = 5/7,
Probability that second ball is not blue = 4/6 = 2/3,
Then:
P(no blue in both draws) = 5/7 × 4/6 = 20/42 = 10/21.
(4)
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)
Malik Ahamed said:
1 year ago
Where did 6 came from? Please explain.
(3)
Harry said:
7 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.
(2)
Tharini said:
6 years ago
There is;
2 Red.
3 Green.
2 Blue Balls.
i.e. 5 Non-blue balls and 2 Blue.
1st ball can be picked from those 5 non-blue balls so that none of them is blue.
Hence n(E)=5 & n(S)= 7.
Hence probability is 5/7.
The next ball will be picked from remaining 6 balls and 4 Non-Blue balls(Since one is already picked).
Hence Probability(2nd picked ball not blue) = 4/6.
Hence,
Total probability = (5/7) * (4/6) = 10/21.
2 Red.
3 Green.
2 Blue Balls.
i.e. 5 Non-blue balls and 2 Blue.
1st ball can be picked from those 5 non-blue balls so that none of them is blue.
Hence n(E)=5 & n(S)= 7.
Hence probability is 5/7.
The next ball will be picked from remaining 6 balls and 4 Non-Blue balls(Since one is already picked).
Hence Probability(2nd picked ball not blue) = 4/6.
Hence,
Total probability = (5/7) * (4/6) = 10/21.
(2)
Naseem Ullah Khan said:
5 years ago
Simple explanation, well done, thanks @Krish.
(2)
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)
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