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 11 of 12.
Madhu said:
1 decade ago
Becuase, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
can u please explain how 2 comes here?
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
can u please explain how 2 comes here?
Pranav said:
1 decade ago
A bag contains 3 red, 2 green and 5 blue balls. 4 balls are drawn at random. What is the probability that 1 green ball?
Neha said:
1 decade ago
What is 3c2, 2c2 etc? could anyone please explain this whole sum according to a class 7 student.
Jay said:
1 decade ago
(7 x 6)
(2 x 1)
Please explan it.
(2 x 1)
Please explan it.
Dileep said:
1 decade ago
The way akash did is exact. It takes into account all the probabilities
Srija said:
1 decade ago
You are right akash!
Shiba said:
1 decade ago
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls. ?
My question is why are you finding it only from 5 balls why not 7 balls.
My question is why are you finding it only from 5 balls why not 7 balls.
Akash said:
1 decade ago
Can we do it by this way?
3c2+2c2+3c1 2c1
---------------
7c2.
I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green. Isn't the solution correct?
3c2+2c2+3c1 2c1
---------------
7c2.
I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green. Isn't the solution correct?
Ekta said:
1 decade ago
Please clarify it which one having best answer of the solution.
Subhash said:
1 decade ago
What Kk585 said is not correct.
Becuase, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
Therefore total is 11/21.
Hence ans is 1-(11/21) is 10/21.
Becuase, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
Therefore total is 11/21.
Hence ans is 1-(11/21) is 10/21.
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