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.
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.
Poojitha said:
4 years ago
Total balls are 15.
Only we have to select red balls in total balls red are 5.
One selected from a total 5/13.
Next, another selected (we have already selected one so 1 ball is reduced from both red balls and total balls) == 4/14.
Here, 2 balls are reduced 1 from one selected ball another from again selected ball so 3/13
5/15 * 4/14* 3/13 = 2/91.
Only we have to select red balls in total balls red are 5.
One selected from a total 5/13.
Next, another selected (we have already selected one so 1 ball is reduced from both red balls and total balls) == 4/14.
Here, 2 balls are reduced 1 from one selected ball another from again selected ball so 3/13
5/15 * 4/14* 3/13 = 2/91.
(1)
Laxman said:
5 years ago
@USHA.
For example, there are 3 red balls and you have to select two balls out of it then possible ways are,
6 ways let me show;
R1,R2 R1,R3 R2,R1 R2,R3 R3,R1 R3,R1.
In these selections, we have R1, R2 and R2, R1 same like the way we consider only one so that we have 3 choices but don't consider you are selecting 3 out of 3 balls.
For example, there are 3 red balls and you have to select two balls out of it then possible ways are,
6 ways let me show;
R1,R2 R1,R3 R2,R1 R2,R3 R3,R1 R3,R1.
In these selections, we have R1, R2 and R2, R1 same like the way we consider only one so that we have 3 choices but don't consider you are selecting 3 out of 3 balls.
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.
Deki Zangmo said:
5 years ago
Total ball = 15.
Out of these, you have to select three balls.
15C3= 455,. We can select 3 balls in 455 ways.
But there is a condition that all 3 balls should be red so,
Out of 5 red balls we have to take out 3;
5C3 = 10,
As a result:
10/455 = 2/91.
Out of these, you have to select three balls.
15C3= 455,. We can select 3 balls in 455 ways.
But there is a condition that all 3 balls should be red so,
Out of 5 red balls we have to take out 3;
5C3 = 10,
As a result:
10/455 = 2/91.
(3)
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?
Akanksha said:
6 years ago
A box contains 6000 cards numbered 1 to 6000. One card is drawn at random from the box. Find the probability that the number is neither divisible by 4 nor 6? Please give me a solution.
(1)
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.
Seenivasan said:
11 months ago
5c3 = 5 * 4 * 3/1 * 2 * 3.
Take 3 commons in the above scenario so we can calculate the sum as 5c2.
Take 3 commons in the above scenario so we can calculate the sum as 5c2.
(3)
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