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 6 of 12.
SALONI said:
10 years ago
How we can know that we need to use ''C'' formula in sum ?
Gmbvbgmkgh said:
10 years ago
Why is E used?
Shanel said:
10 years ago
Hey it should be 5/7.
Because if we subtract no of blue from total.
Then 7-(2+3) log%5 = 7*5*3*5*6 = log 7+5+3+6+1 = log 35 + log 15 + log 30.
So if we take log as common the it will be easy i.e = log (35+15+30).
= log*80 = 80 log the 80-75 = 5.
There fore the possibility is 5/7.
Because if we subtract no of blue from total.
Then 7-(2+3) log%5 = 7*5*3*5*6 = log 7+5+3+6+1 = log 35 + log 15 + log 30.
So if we take log as common the it will be easy i.e = log (35+15+30).
= log*80 = 80 log the 80-75 = 5.
There fore the possibility is 5/7.
Kenneth said:
10 years ago
Please clear me on how 6 come in?
Faffy said:
9 years ago
What if we use the tree diagrams because the way you answer is kindly complicated.
Rizwan said:
9 years ago
Why not the answer is 5/7?
Burhan said:
9 years ago
Could anybody tell me what is that c means in 7c2?
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)
Roma said:
9 years ago
You are right Saurabh, in my way the method 1 is correct.
Amor said:
9 years ago
Agree with you @Roma.
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