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 5 of 12.
Anil pradhan said:
1 decade ago
Simple and clear way is:
Probability of getting blue is 2/7.
And getting non-blue is 1-2/7 = 5/7.
i.e Total probability-Probability of blue ball.
Probability of getting blue is 2/7.
And getting non-blue is 1-2/7 = 5/7.
i.e Total probability-Probability of blue ball.
Prabhat kumar mishra said:
1 decade ago
Total number of balls = 7.
Therefore first probability is = 5/7and second probability will be 4/6.
Hence, total probability = (5/7)*(4/6) = 20/42 = 10/21.
Therefore first probability is = 5/7and second probability will be 4/6.
Hence, total probability = (5/7)*(4/6) = 20/42 = 10/21.
ARUN KASHYAP said:
1 decade ago
Answer is simple:
2+3+2 = 7x3(here 3 is no.of 2+3+2) = 21.
2+ 3(here 3 is no.of 2+3+2) = 5x2 = 10.
10/21.
2+3+2 = 7x3(here 3 is no.of 2+3+2) = 21.
2+ 3(here 3 is no.of 2+3+2) = 5x2 = 10.
10/21.
Mohammad Javed said:
1 decade ago
There is one easy way.
No balls should be blue means either it should be red or green which sums to 5 balls. Total balls is 7.
Probability of no blue will be = (5/7)*(4/6)=10/21.
As we have picked one ball we are reducing Red+Green balls to 4. And total balls to 6.
No balls should be blue means either it should be red or green which sums to 5 balls. Total balls is 7.
Probability of no blue will be = (5/7)*(4/6)=10/21.
As we have picked one ball we are reducing Red+Green balls to 4. And total balls to 6.
Stephen said:
1 decade ago
Is there an easier way of doing these type of probability word problems?
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:
10 years 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:
10 years 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?
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