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 6 of 6.
Naresh said:
1 decade ago
"And" is asked in problem -> Multiplication.
"Or" is asked in problem -> Addition.
"Or" is asked in problem -> Addition.
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.
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!
Chandni said:
10 years ago
Can anybody tell me why are they using 25C3?
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