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:
119 comments Page 12 of 12.
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:
2 decades ago
Please clarify it which one having best answer of the solution.
Subhash said:
2 decades 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.
Rose were said:
2 decades ago
How did the multiplications come about? please answer me.
Rose were said:
2 decades ago
Why are you multiplying by 6 and by 4?
Saurabh said:
2 decades ago
This is the the simplest method to find the Probablity...........
Kk585 said:
2 decades ago
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 none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Soundar said:
2 decades ago
Its a correct answer and short way.
Aarthi said:
2 decades ago
Can we do this n different method? finding the possibility that both are blue and finally ans-1.
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