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 3 of 4.
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.
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?
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.
Dharika said:
1 decade ago
White = 4C3.
So white = 4.
Red = 5C3.
So red = 10.
Blue = 6C3.
So blue = 20.
Now total is 34 and we want red so 34-24 =10.
And total cases are 15C3. So answer will be 10/455 = 2/91.
Can we do it like this?
So white = 4.
Red = 5C3.
So red = 10.
Blue = 6C3.
So blue = 20.
Now total is 34 and we want red so 34-24 =10.
And total cases are 15C3. So answer will be 10/455 = 2/91.
Can we do it like this?
Narasimha said:
1 decade ago
Given Qns, tree balls are drawn randomly out of 4 white, 5 red, 6 blue balls, ok.
Qns asked what is the probability of getting all are red balls?
Event is all are red balls n(E) = 5C3.
Total outcomes is n(S) = 15C3.
P(E) = n(E)/n(S) = 5C3/15C3.
Qns asked what is the probability of getting all are red balls?
Event is all are red balls n(E) = 5C3.
Total outcomes is n(S) = 15C3.
P(E) = n(E)/n(S) = 5C3/15C3.
Priya said:
1 decade ago
Can anyone explain clearly?
Dhruvil said:
1 decade ago
Probability = (5*4*3)/(15*14*13) = (6/273) = (2/91).
As there are 5 red balls in each bag.
As there are 5 red balls in each bag.
Rohit said:
1 decade ago
1st red ball 5/15.
2nd red ball 4/14.
3rd red ball 3/13.
5*4*3/15*14*13 = 2/91.
2nd red ball 4/14.
3rd red ball 3/13.
5*4*3/15*14*13 = 2/91.
USHA said:
1 decade ago
All of them red is possible only in one way so event is E= 5C1.
But how come 5c3 ?
But how come 5c3 ?
Naushad Farooqi said:
1 decade ago
For example: 5C3 = 5C2
5C3 = 5*4*3/3*2*1 = 5*4/2*1 (here 3 get cancelled)
5C2 = 5*4/2*1
Therefore, 5C3 = 5C2.
In 5C3, we can subtract 3 from 5 i.e. 5-3=2
So 5C3 will be equal to 5C2
There is question what is the use of it ?
to minimize the calculation .i.e. in case of 5C2 there are less calculation compare to 5C3.
Similarly: 6C4=6C2
Hope this will help you.
5C3 = 5*4*3/3*2*1 = 5*4/2*1 (here 3 get cancelled)
5C2 = 5*4/2*1
Therefore, 5C3 = 5C2.
In 5C3, we can subtract 3 from 5 i.e. 5-3=2
So 5C3 will be equal to 5C2
There is question what is the use of it ?
to minimize the calculation .i.e. in case of 5C2 there are less calculation compare to 5C3.
Similarly: 6C4=6C2
Hope this will help you.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers