Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 12)
12.
A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Answer: Option
Explanation:
Let S be the sample space.
| Then, n(S) | = number of ways of drawing 3 balls out of 15 | |||
| = 15C3 | ||||
|
||||
| = 455. |
Let E = event of getting all the 3 red balls.
 n(E) = 5C3 = 5C2 = |
(5 x 4) | = 10. |
| (2 x 1) |
| P(E) = |
n(E) | = | 10 | = | 2 | . |
| n(S) | 455 | 91 |
Discussion:
33 comments Page 2 of 4.
Harsh said:
10 years ago
@Dharika.
Blue and white not to be calculated. Why doing extra efforts and wasting time?
Simply do for red and total cases. That's it.
Blue and white not to be calculated. Why doing extra efforts and wasting time?
Simply do for red and total cases. That's it.
Keisha said:
9 years ago
S = 15.
e = 5/15 (red balls/total balls).
= 1/3.
Why is this not the answer?
e = 5/15 (red balls/total balls).
= 1/3.
Why is this not the answer?
M SAIREDDY said:
9 years ago
Here n(s) = 15;
In question how many balls are random = 3 balls so i.e. = 15C3 = (15*14*13/3*2*1) = 455 = n(s).
In about they find out the only probability getting the only red ball.
So total no.of red balls = 5.
In 3 balls random, so we get = 5C3 = (5*4*3/3*2*1) = 10 = n(e).
In probability = n(e)/n(s) = 10/455 = 2/91.
In question how many balls are random = 3 balls so i.e. = 15C3 = (15*14*13/3*2*1) = 455 = n(s).
In about they find out the only probability getting the only red ball.
So total no.of red balls = 5.
In 3 balls random, so we get = 5C3 = (5*4*3/3*2*1) = 10 = n(e).
In probability = n(e)/n(s) = 10/455 = 2/91.
Hiren said:
9 years ago
I like your logic @Dharika.
Alab said:
9 years ago
4 + 5 + 6 = 15.
First draw is 5/15.
Second 4/14.
Third 3/13.
(5/15)(4/14)(3/13) = 2/91
First draw is 5/15.
Second 4/14.
Third 3/13.
(5/15)(4/14)(3/13) = 2/91
(1)
Pandu raju said:
9 years ago
Thank you for the explanation.
Chetan said:
8 years ago
Good explanation, Thanks @Rohit.
Adu said:
8 years ago
Thanks all.
Somya said:
8 years ago
While calculating the term for red balls, why did we not consider the blue balls?
Rahul. Moorkoth said:
7 years ago
According to me, the Answer should be 4/91.
(1)
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