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 12 of 12.
Riafat Hussain said:
1 year ago
1st ball = 5/7.
2nd ball = 4/6.
5/7 * 4/6 = 10/21.
2nd ball = 4/6.
5/7 * 4/6 = 10/21.
(61)
Sneha said:
1 year ago
Can anyone explain this in the short method?
(8)
Malik Ahamed said:
1 year ago
Where did 6 came from? Please explain.
(3)
Abhikant said:
8 months ago
Two balls removed at random,- these two balls removed one by one or at once it is not mentioned.
If it's mentioned one by one then the answer would be 10/21.
But both balls and none of the balls drawn are blue then probability = 1 - 2/7 = 5/7.
If it's mentioned one by one then the answer would be 10/21.
But both balls and none of the balls drawn are blue then probability = 1 - 2/7 = 5/7.
(10)
Priya said:
5 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.
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)
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