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 4 of 12.
Pooja said:
9 years ago
Please solve this question:
A bag contains 5 white, 6 red, 2 green and 2 black balls. One ball is selected at random from the bag. Find the probability that the selected ball is;
A) white
B) non-white
C) white or green
D) black or red
A bag contains 5 white, 6 red, 2 green and 2 black balls. One ball is selected at random from the bag. Find the probability that the selected ball is;
A) white
B) non-white
C) white or green
D) black or red
John said:
1 decade ago
Probability of occurrence of blue ball.
1 blue AND 1 blue OR,
1 red/green AND 1 blue OR,
1 blue AND 1 red/green.
(2/7 * 1/6) + (5/7 * 2/6) + (2/7 * 5/6) = 11/21.
Probability of non-occurrence of blue ball = 1-(11/21) = (10/21) ANS.
1 blue AND 1 blue OR,
1 red/green AND 1 blue OR,
1 blue AND 1 red/green.
(2/7 * 1/6) + (5/7 * 2/6) + (2/7 * 5/6) = 11/21.
Probability of non-occurrence of blue ball = 1-(11/21) = (10/21) ANS.
Manisha said:
9 years ago
Please solve this question:
A bank contains 5white,6red,2green, and 2black balls one ball is selected at random from the bag. Find the probability that the selected ball is:
1. White
2. Non white
3. White or green
4. Black or red
A bank contains 5white,6red,2green, and 2black balls one ball is selected at random from the bag. Find the probability that the selected ball is:
1. White
2. Non white
3. White or green
4. Black or red
ABI said:
5 years ago
The answer should be 5/7 right!
Because Probability = No.of favourable outcomes/Total outcomes.
=5/7.
Since, if both balls r not blue then they have to be either red(2) or green (3).
Thus,2+3 is 5.
So, option D is correct.
Because Probability = No.of favourable outcomes/Total outcomes.
=5/7.
Since, if both balls r not blue then they have to be either red(2) or green (3).
Thus,2+3 is 5.
So, option D is correct.
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.
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.
Thejo said:
5 years ago
We should pick 2 from 7 none of the ball drawn is blue, then the picked balls colours should be red and green. How come it is 10/21?
Here it could be 2C1*3C1 ÷ 7C2 = 6/21 right! Can any1 tell me why it is wrong?
Here it could be 2C1*3C1 ÷ 7C2 = 6/21 right! Can any1 tell me why it is wrong?
(2)
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).
Shashikant sahu said:
7 years ago
This is a very simple technique.
1st ball not to be blue is = (1-2/7)=5/7.
Now next ball not to be blue is = (1-2/6)=4/6.
*(causes 1 ball has already taken).
Multiple both;
= 5/7 * 4/6.
= 10/21.
1st ball not to be blue is = (1-2/7)=5/7.
Now next ball not to be blue is = (1-2/6)=4/6.
*(causes 1 ball has already taken).
Multiple both;
= 5/7 * 4/6.
= 10/21.
IMx said:
1 month ago
If we did it your way, assuming:
Probability that the first ball is not blue = 5/7,
Probability that second ball is not blue = 4/6 = 2/3,
Then:
P(no blue in both draws) = 5/7 × 4/6 = 20/42 = 10/21.
Probability that the first ball is not blue = 5/7,
Probability that second ball is not blue = 4/6 = 2/3,
Then:
P(no blue in both draws) = 5/7 × 4/6 = 20/42 = 10/21.
(5)
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