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 8 of 12.
Chandani said:
1 decade ago
Please answer this question. What is the number of ways of selecting 3 balls from a bag containing 5 blue and 6 red balls. If the answer is 11C3 please explain what about the cases where 2 are blue and 1 red.
This will happen a number of times, but they will all be counted as separate selections. How do we account for the similarity. Please help.
This will happen a number of times, but they will all be counted as separate selections. How do we account for the similarity. Please help.
Sathish kumar said:
1 decade ago
Total 7 balls.
Blue 2 balls.
So none of blue is 5 balls : 5c2/7c2.
(5x4/2x1)/(7x6/2x1) = 10/21 answer.
Blue 2 balls.
So none of blue is 5 balls : 5c2/7c2.
(5x4/2x1)/(7x6/2x1) = 10/21 answer.
Itachi said:
1 decade ago
Show the answer using conditional probability please!
Zishan said:
1 decade ago
There are two bags A and B. A contains n white and 2 black balls and B contains 2 white and n black balls. One of the two bags is selected at random and two balls are drawn from it without replacement.
If both the balls drawn are white and the probability that the bag A was used to draw the balls is 6/7, find the value of n.
Please answer with solution.
If both the balls drawn are white and the probability that the bag A was used to draw the balls is 6/7, find the value of n.
Please answer with solution.
Jitesh mittal said:
1 decade ago
Cannot we have a different method.
It will be like 1- putting out blue balls which is 2/7.
To probability will be 1-2/7 = 5/7?
It will be like 1- putting out blue balls which is 2/7.
To probability will be 1-2/7 = 5/7?
SALONI said:
1 decade ago
How we can know that we need to use ''C'' formula in sum ?
Gmbvbgmkgh said:
1 decade 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:
10 years ago
What if we use the tree diagrams because the way you answer is kindly complicated.
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