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 2 of 6.
Abhi said:
1 decade ago
I solved this problem as
(10/25)*(15/24)*(14/23)
because 1st select 1 girl from 10 out of total 25 students, then 1 boy from 15 out of remaining 24 then again 1 boy from 14 boys and total remainig 23 students.... i got 7/46 as my ans.....
Can anyone please tell me whats wrong? please..
(10/25)*(15/24)*(14/23)
because 1st select 1 girl from 10 out of total 25 students, then 1 boy from 15 out of remaining 24 then again 1 boy from 14 boys and total remainig 23 students.... i got 7/46 as my ans.....
Can anyone please tell me whats wrong? please..
Jahid said:
9 years ago
@Nemi,
The equation for solving your problem should be as follows:
Let assume total women in the group is x.
The probability of being a 1st choice woman is x/(x + 5).
2nd choice woman probability is (x - 1)/(x + 5 - 1).
So equation will be,
x/(x + 5) * x - 1/(x + 4) = 3/20.
The equation for solving your problem should be as follows:
Let assume total women in the group is x.
The probability of being a 1st choice woman is x/(x + 5).
2nd choice woman probability is (x - 1)/(x + 5 - 1).
So equation will be,
x/(x + 5) * x - 1/(x + 4) = 3/20.
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?
Annie said:
1 decade ago
@Abhi
Because we do not care about the order in which we pick the boys and girls.
Also, if you picked the boys and girls in a different order than the one you used, wouldn't it be a different answer? So we don't solve it like x/25 x y/24 x z/23
Because we do not care about the order in which we pick the boys and girls.
Also, if you picked the boys and girls in a different order than the one you used, wouldn't it be a different answer? So we don't solve it like x/25 x y/24 x z/23
Usman said:
1 decade ago
It is very simple in the probability theorem and means *(multiplication)n or means +(addition).
So in the above question they asked to find out probability of selected students are one girl and two boy.
So 10c1*15c2/25c3 = 1050/2300 = 21|46.
So in the above question they asked to find out probability of selected students are one girl and two boy.
So 10c1*15c2/25c3 = 1050/2300 = 21|46.
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).
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!
Shyam said:
1 decade ago
@Abhi.
I have got the solution for our problem. We are computing the probability of just one combination. i.e we need to do P(1g,1b,1b)+P(1b,1g,1b)+P(1b,1b,1g). we found only the first combination. :) Hope you understood.
I have got the solution for our problem. We are computing the probability of just one combination. i.e we need to do P(1g,1b,1b)+P(1b,1g,1b)+P(1b,1b,1g). we found only the first combination. :) Hope you understood.
Kruvy said:
1 decade ago
You got the Point @Abhi .. :-)
However I cannot help you in this. As I have no clue about is this a right method and robust to solve same kind of problem...??!
Help us.
And I Like this site. Appreciate it @IndiaBix.
However I cannot help you in this. As I have no clue about is this a right method and robust to solve same kind of problem...??!
Help us.
And I Like this site. Appreciate it @IndiaBix.
Karthick M said:
3 years ago
@Prabhu and @Priyaa.
The question is given as 1 girl and 2 boys. Given 'AND' we use simple of multiplication.
Suppose they are given In the question if 1 girl or 2 girls. We use simple subtraction.
That's all.
The question is given as 1 girl and 2 boys. Given 'AND' we use simple of multiplication.
Suppose they are given In the question if 1 girl or 2 girls. We use simple subtraction.
That's all.
(8)
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