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.
Srija said:
1 decade ago
You are right akash!
Dileep said:
1 decade ago
The way akash did is exact. It takes into account all the probabilities
Jay said:
1 decade ago
(7 x 6)
(2 x 1)
Please explan it.
(2 x 1)
Please explan it.
Neha said:
1 decade ago
What is 3c2, 2c2 etc? could anyone please explain this whole sum according to a class 7 student.
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?
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?
Nishant said:
1 decade ago
Can you please explain me why we select Number of ways of drawing 2 balls out of (2 + 3) balls instead of there having total no.of 7 balls, . So prob. Of drawing 2 balls must be out of 7 i.e. (2+3+2).
Krish said:
1 decade ago
Because, 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?.""""
But the question says NO BLUE BALL..why are we considering 1 blue and 1 green ...or 1 redor 1 blue..?
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?.""""
But the question says NO BLUE BALL..why are we considering 1 blue and 1 green ...or 1 redor 1 blue..?
Krish said:
1 decade ago
I am not geting wats wrong in this...please explain....
Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21
Probability that none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21
Probability that none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Jaani! said:
1 decade ago
@krish, Cuz 2C2 means picking those 2 blue balls out of those 2 blue balls only! Bt d ques says "None of them blue" bt one blue and one of any other colour cant b included also! Thats y we do that of picking up only green and red.. (no possiblty of blue)!
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