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 1 of 4.
Divya said:
1 decade ago
How come 5C3 become 5C2? please explain.
Dinesh guptha said:
1 decade ago
Don't confuse divya.
Keep it as 5C3.
You will get the same ans.
Cool!
Keep it as 5C3.
You will get the same ans.
Cool!
Rohit jadhav said:
1 decade ago
Ya dinesh is right.
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.
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 ?
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.
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.
Priya said:
1 decade ago
Can anyone explain clearly?
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.
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?
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