Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 7)
7.
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
Answer: Option
Explanation:
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
| Then, n(S) | = Number ways of selecting 3 students out of 25 | |||
| = 25C3 ` | ||||
|
||||
| = 2300. |
| n(E) | = (10C1 x 15C2) | ||||||
|
|||||||
| = 1050. |
 P(E) = |
n(E) | = | 1050 | = | 21 | . |
| n(S) | 2300 | 46 |
Discussion:
54 comments Page 3 of 6.
Archana Mahour said:
9 years ago
Please provide a solution for this question.
A class consists of 80 students, 25 of them are girls and 55 boys, 10 of them are rich and the remaining poor, 20 of them are fair complexion. What is the probability of selecting a fair complexioned rich girl?
A class consists of 80 students, 25 of them are girls and 55 boys, 10 of them are rich and the remaining poor, 20 of them are fair complexion. What is the probability of selecting a fair complexioned rich girl?
Person said:
9 years ago
There are three probabilities, BBG, BGB and GBB. The chance of getting 2 boys and 1 girl is 15/25 X 14/24 X 10/23 which is 7/46. Since there are three ways to get 2 boys and a Girl,
7/46 X 3 = 21/46.
7/46 X 3 = 21/46.
(2)
Anony said:
9 years ago
Why is it not solved like this?
Total possible combinations - BBB, GGG, BGB, BBG, GBB, GGB, GBG, BGG i.e. 8.
Fav combination: GBB, BGB, BBG so 3.
So the answer should be 3/8.
Total possible combinations - BBB, GGG, BGB, BBG, GBB, GGB, GBG, BGG i.e. 8.
Fav combination: GBB, BGB, BBG so 3.
So the answer should be 3/8.
Kathir lee said:
9 years ago
@ Sanjana.
Thank you for the explanation.
Thank you for the explanation.
Azly said:
9 years ago
Thank you! @Sanjana.
Naresh rajpurohit said:
10 years ago
25 students and 3 select boys and girl.
So number ways of selecting 3 students out of 25. So use 25C3.
And Probability = 10C1 (10 girl in choose any one) * 15C2 (15 boys in choose any two)/25C3 (All students in select 1 girl and 2 boys).
So number ways of selecting 3 students out of 25. So use 25C3.
And Probability = 10C1 (10 girl in choose any one) * 15C2 (15 boys in choose any two)/25C3 (All students in select 1 girl and 2 boys).
Mary-ann said:
10 years ago
I can't understand someone help.
Chandni said:
10 years ago
Can anybody tell me why are they using 25C3?
John said:
1 decade ago
Whether you choose BBG, BGB or GBB you have the same numerators of 15, 14 and 10 AND the same denominators of 25, 24 and 23.
That works out as (15x14x10 = 2100) and (25x24x23 = 13800) which factors down to 7/46. I think I hope!
That works out as (15x14x10 = 2100) and (25x24x23 = 13800) which factors down to 7/46. I think I hope!
Asc said:
1 decade ago
@Abhi.
You got this wrong when you took only one girl as the first solution. You could also have one boy and another boy as the first student chosen i.e 1.
You got this wrong when you took only one girl as the first solution. You could also have one boy and another boy as the first student chosen i.e 1.
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