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.
RAJA said:
1 decade ago
Very good sanjana.
Superprofessor Extraordinaire said:
1 decade ago
The answer above is wrong. There are 3 ways to get 1 girl and 2 boys when choosing 3: (1)BBG, (2)BGB, (3)GBB. Probability of observing outcome (1) is 15/25*15/25*10/25=.6*.6*.4=.144.The probability of observing outcome (2) is 15/25*10/25*1/25=.6*.4*.6=.144. The probability of observing outcome (3) is 10/25*10/25*15/25=.4*.6*.6=.144.
Sum the three possible outcomes to get the probablity of the event .144+.144+.144=.432.
This is just a binomial probability where the probability of a success, say choosing a boy is 15/25=.6 and we want to know the probability of getting 2 boys in 3 chance, which is .432.
Sum the three possible outcomes to get the probablity of the event .144+.144+.144=.432.
This is just a binomial probability where the probability of a success, say choosing a boy is 15/25=.6 and we want to know the probability of getting 2 boys in 3 chance, which is .432.
Shyam said:
1 decade ago
@Abhi.
I have the same problem. I am getting 7/46 But by formula the answer is given as 21/46. I solved it by (10/25) * (15/24) * (14/23).
Please anyone explain on this query.
I have the same problem. I am getting 7/46 But by formula the answer is given as 21/46. I solved it by (10/25) * (15/24) * (14/23).
Please anyone explain on this query.
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.
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
Ankit Patel said:
1 decade ago
@Abhi we can do it by this way.
For girl : 10/25 (total students 10 + 15 = 25).
For 1st boy: 15/24 (1 girl is out so total is 24).
For 2nd boy: 14/23 (now total boys 14 and total students are 23).
[(10*15*14)/(25*24*23)]
=7/46.
But here three possibility so multiply by 3
1st 1g 1b 1b
2nd 1b 1g 1b
3rd 1b 1b 1g
So, answer = 21/46.
For girl : 10/25 (total students 10 + 15 = 25).
For 1st boy: 15/24 (1 girl is out so total is 24).
For 2nd boy: 14/23 (now total boys 14 and total students are 23).
[(10*15*14)/(25*24*23)]
=7/46.
But here three possibility so multiply by 3
1st 1g 1b 1b
2nd 1b 1g 1b
3rd 1b 1b 1g
So, answer = 21/46.
Nemi said:
1 decade ago
If there are 5 men and some women in a group. If the probability to select two Girls out of all is 3/20 then find the number of Girls in a group. Give answer if any one know.
Ajay said:
1 decade ago
We can't take (10/25) (15/24) (14/23). , because all 3 students are selected at once, not one after another.
Jay lee lord said:
1 decade ago
The answer is correct.
I have alternative solution.
There are 3 possible ways in picking 1 girl and 2 boys.
Girl , boy , boy
Boy , girl , boy
Boy , boy , girl
Gbb:
(10/25)*(15/24)*(14/23)=7/46.
Bgb:
(15/25)*(10/24)*(14/23)=7/46.
Bbg:
(15/25)*(14/24)*(10/23)=7/46.
Adding them all will give you 21/46.
I have alternative solution.
There are 3 possible ways in picking 1 girl and 2 boys.
Girl , boy , boy
Boy , girl , boy
Boy , boy , girl
Gbb:
(10/25)*(15/24)*(14/23)=7/46.
Bgb:
(15/25)*(10/24)*(14/23)=7/46.
Bbg:
(15/25)*(14/24)*(10/23)=7/46.
Adding them all will give you 21/46.
Dhruvil said:
1 decade ago
Probability = (15*14+10/25*24*23).
= (220/25*24*23) = (11*2*10/25*24*23) = (11/5*6*23) = (11/690).
= (220/25*24*23) = (11*2*10/25*24*23) = (11/5*6*23) = (11/690).
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers