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 2 of 12.
Sagar said:
1 decade ago
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 `
= (7 x 6) /(2 x 1)
= 21.
A: Event none of the balls draw is blue.
n(A): number of possible ways = 2C1*3C1+2C2*3C0+3C2*2C0
=2*3+1*1+3*1
=6+1+3
=10.
and
P(A)=n(A)/n(S)=10/21=1/2.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2 `
= (7 x 6) /(2 x 1)
= 21.
A: Event none of the balls draw is blue.
n(A): number of possible ways = 2C1*3C1+2C2*3C0+3C2*2C0
=2*3+1*1+3*1
=6+1+3
=10.
and
P(A)=n(A)/n(S)=10/21=1/2.
Prabhat said:
1 decade ago
There is
2 Red
3 Green
2 Blue Balls
i.e. 5 Non blue balls and 2 Blue
1st ball can be picked from those 5 non blue balls so that none of them is blue.
Hence n(E)=5 & n(S)= 7
Hence probability is 5/7
The next ball will be picked from remaining 6 balls and 4 Non-Blue balls(Since one is already picked)
Hence Probability(2nd picked ball not blue) = 4/6
Hence
Total probability = (5/7) * (4/6) = 10/21.
2 Red
3 Green
2 Blue Balls
i.e. 5 Non blue balls and 2 Blue
1st ball can be picked from those 5 non blue balls so that none of them is blue.
Hence n(E)=5 & n(S)= 7
Hence probability is 5/7
The next ball will be picked from remaining 6 balls and 4 Non-Blue balls(Since one is already picked)
Hence Probability(2nd picked ball not blue) = 4/6
Hence
Total probability = (5/7) * (4/6) = 10/21.
Subhash said:
1 decade ago
What Kk585 said is not correct.
Becuase, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
Therefore total is 11/21.
Hence ans is 1-(11/21) is 10/21.
Becuase, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
Therefore total is 11/21.
Hence ans is 1-(11/21) is 10/21.
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.
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.
Saiteja said:
2 years ago
A bag contains 2 red, 3 green and 2 blue balls, total = 7 balls.
(probability of choosing+probability of not choosing=1).
Let's consider the probability of choosing two balls is blue then,
probability of choosing P=(2/7)*(1/6) = 1/21.
so, the probability of not choosing (NP) = 1-P,
NP = 1-(1/21)
NP = 20/21,
I think 20/21 is the right answer.
(probability of choosing+probability of not choosing=1).
Let's consider the probability of choosing two balls is blue then,
probability of choosing P=(2/7)*(1/6) = 1/21.
so, the probability of not choosing (NP) = 1-P,
NP = 1-(1/21)
NP = 20/21,
I think 20/21 is the right answer.
(27)
Anhad said:
2 years ago
@Saiteja
I also thought that 20/21 is correct, actually, you have only deducted the probability of both blue from 1, you have to also deduct at least one blue from 1 i.e. 10/21. So the final calculation will go 1 - 1/21 - 10/21 = 10/21. (Probability of not getting blue = 1 - the probability of both blue - probability of one blue)
I also thought that 20/21 is correct, actually, you have only deducted the probability of both blue from 1, you have to also deduct at least one blue from 1 i.e. 10/21. So the final calculation will go 1 - 1/21 - 10/21 = 10/21. (Probability of not getting blue = 1 - the probability of both blue - probability of one blue)
(12)
Madhu said:
1 decade ago
Becuase, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
can u please explain how 2 comes here?
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
can u please explain how 2 comes here?
Krish said:
5 years ago
Using conceptual method rather than going by formula gives a better feel for what is happening.
P1 = probability of 1st ball not blue = 5/7 as 5 are non-blue.
P2 = probability of 2nd ball not blue = 4/6 as 1 non-blue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.
P1 = probability of 1st ball not blue = 5/7 as 5 are non-blue.
P2 = probability of 2nd ball not blue = 4/6 as 1 non-blue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.
(10)
Farah said:
1 decade ago
Formula is probability is p(A) = n(A)/n(s), so we find n(s) = 7C2 = 21.
Next we find n(a) = 5C2 = 10 these amounts put in formula
p(A)=n(A)/n(S) Ans is 10/21
We have interested in 2 balls none in white so 2+5 not 2 white balls I think you understood
Next we find n(a) = 5C2 = 10 these amounts put in formula
p(A)=n(A)/n(S) Ans is 10/21
We have interested in 2 balls none in white so 2+5 not 2 white balls I think you understood
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