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 5 of 6.
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.
Prabhu said:
1 decade ago
(10C1 x 15C2) why they are multiplying this ?
They should add the value know .. 10 c1 + 15 c2
They should add the value know .. 10 c1 + 15 c2
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).
Priya said:
1 decade ago
1050/2300 = 21/46.
Yugesh said:
1 decade ago
Why do you multiply 24x25x23?
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.
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