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 3 of 12.
Thejo said:
5 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 ÷ 7C2 = 6/21 right! Can any1 tell me why it is wrong?
Here it could be 2C1*3C1 ÷ 7C2 = 6/21 right! Can any1 tell me why it is wrong?
(2)
Somesh Saurabh said:
9 years ago
See there are two ways to solve this question:
METHOD 1:
Probability of getting no blue balls in 2 draws = probability of getting 2 other balls in 2 draws out of any other 5 balls.
i.e., select two balls from the remaining 5 balls in 5C2 ways = n(E),
Now the sample space is n(S) = selection of 2 balls out 7b balls = 7C2,
So, required probability= n(E)/n(S),
=> 5C2/7C2 = [5!/(2!*3!)]/[7!/(2!*5!)]
=> 10/21 is the answer.
METHOD 2:
Probability of not getting any blue balls in two draws = 1 - (probability of getting 1 blue ball and 1 other ball + probability of getting 2 blue balls)
So,
Selection of 1 blue ball and one other ball = 2C1 * 5C1 = 10 ways,
Selection of 2 blue balls = 2C2 = 1,
So, probability = 1 - [{10/7C2} + {1/7C2}]
=> 10/21 answer.
METHOD 1:
Probability of getting no blue balls in 2 draws = probability of getting 2 other balls in 2 draws out of any other 5 balls.
i.e., select two balls from the remaining 5 balls in 5C2 ways = n(E),
Now the sample space is n(S) = selection of 2 balls out 7b balls = 7C2,
So, required probability= n(E)/n(S),
=> 5C2/7C2 = [5!/(2!*3!)]/[7!/(2!*5!)]
=> 10/21 is the answer.
METHOD 2:
Probability of not getting any blue balls in two draws = 1 - (probability of getting 1 blue ball and 1 other ball + probability of getting 2 blue balls)
So,
Selection of 1 blue ball and one other ball = 2C1 * 5C1 = 10 ways,
Selection of 2 blue balls = 2C2 = 1,
So, probability = 1 - [{10/7C2} + {1/7C2}]
=> 10/21 answer.
(1)
Anand Nithi said:
8 years ago
R2 ,G3, B2.
Total = 2+3+2=7.
probability = 2/7 *3/7 * 2/7 = 12/21.
either R or G ..but not B.
probability of B = 2/21.
So,
(12/21) - (2/21) = 10/21.
Is this correct?
Total = 2+3+2=7.
probability = 2/7 *3/7 * 2/7 = 12/21.
either R or G ..but not B.
probability of B = 2/21.
So,
(12/21) - (2/21) = 10/21.
Is this correct?
(1)
Gufran said:
5 years ago
(2R+3G)c2/7c2.
5c2/7c2.
5 * 2/2/7 * 6/2.
(20/2)/(42/2).
10/21 is the final answer
5c2/7c2.
5 * 2/2/7 * 6/2.
(20/2)/(42/2).
10/21 is the final answer
(1)
Priya said:
8 months ago
(2c0 3c1 + 2c1 3c1 + 2c2 3c0)/7c2.
Which is equal to (3+6+1)/21. = 10/21.
Which is equal to (3+6+1)/21. = 10/21.
(1)
Aarthi said:
2 decades ago
Can we do this n different method? finding the possibility that both are blue and finally ans-1.
Soundar said:
2 decades ago
Its a correct answer and short way.
Kk585 said:
1 decade 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 .......
Saurabh said:
1 decade ago
This is the the simplest method to find the Probablity...........
Rose were said:
1 decade ago
How did the multiplications come about? please answer me.
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