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.
Nishant said:
1 decade ago
Can you please explain me why we select Number of ways of drawing 2 balls out of (2 + 3) balls instead of there having total no.of 7 balls, . So prob. Of drawing 2 balls must be out of 7 i.e. (2+3+2).
Krish said:
1 decade ago
I am not geting wats wrong in this...please explain....
Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21
Probability that none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21
Probability that none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Krish said:
1 decade ago
Because, 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?.""""
But the question says NO BLUE BALL..why are we considering 1 blue and 1 green ...or 1 redor 1 blue..?
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?.""""
But the question says NO BLUE BALL..why are we considering 1 blue and 1 green ...or 1 redor 1 blue..?
Jaani! said:
1 decade ago
@krish, Cuz 2C2 means picking those 2 blue balls out of those 2 blue balls only! Bt d ques says "None of them blue" bt one blue and one of any other colour cant b included also! Thats y we do that of picking up only green and red.. (no possiblty of blue)!
Rajee said:
1 decade ago
Here.
No of out comes is 7c2 and no of events is 3c2+2c2+2c1. 3c1 then no of events possible is.
P (E) =n (E) /n (s) =11/21.
But we have to find here is none of the two ball is blue so we have to find.
P (~E) =1-P (E) =10/21.
No of out comes is 7c2 and no of events is 3c2+2c2+2c1. 3c1 then no of events possible is.
P (E) =n (E) /n (s) =11/21.
But we have to find here is none of the two ball is blue so we have to find.
P (~E) =1-P (E) =10/21.
Piyul said:
1 decade ago
Is there any other way to understand basic concept of probability?
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
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.
Kavita kant said:
1 decade ago
Is there any way to determine that when we take combination to find probability and when we take direct favourable events/total no of events. Actually I am confused in determining where to take combination and where not.
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.
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