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Aptitude - Probability - Discussion

Discussion Forum : Probability - General Questions (Q.No. 7)
Probability
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:
21
46
25
117
1
50
3
25
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 `
= (25 x 24 x 23)
(3 x 2 x 1)
= 2300.

n(E) = (10C1 x 15C2)
= 10 x (15 x 14)
(2 x 1)
= 1050.

P(E) = n(E) = 1050 = 21 .
n(S) 2300 46

Discussion:
54 comments Page 4 of 6.
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).
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.
Prabhu said:   1 decade ago
(10C1 x 15C2) why they are multiplying this ?

They should add the value know .. 10 c1 + 15 c2
Naresh said:   1 decade ago
"And" is asked in problem -> Multiplication.

"Or" is asked in problem -> Addition.
Sree said:   9 years ago
It is simple.

Only using the equation (n* (n-1) )/r!. We will get the solution.
Shekhar Gupta said:   8 years ago
To select boy and girls are mutually independent events. So they are multiplied.
Priya said:   1 decade ago
@prabhu.

Even I have the same doubt. Can some intellectual person answer this?
Shikha said:   8 years ago
I am not able to understand this answer please some one clear my doubt.
Ronald said:   3 years ago
I don't understand the answer, please someone help me.
(7)
Gnana said:   1 decade ago
I too have same doubt. Can any one clarify this, soon.
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