Probability

Question

Given 4 children lives in Mrs. A home. One fine day Mrs. A reveals to Mr. A that atleast 1 out of 4 children is a girl. Probability that there are exactly 3 girl in Mrs. A's home __________

Comments

it should be 1/4

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4/15??

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0.125?
As 1 is girl atleast. Rest 3 can be boys or girls.
Thus, total possibilities = 1 x 2 x 2 x 2 = 8

Case : 3 are girls. So 4th has to be guy. Hence 1 x 1 x 1 x 1=1

1/8 = 0.125

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let E_3G ->exactly 3 girls, At_1G -> atleast 1 girl

total possible combinations of (#boys, #girls) = (0,4), (1,3), (2,2), (3,1), (4,0)

P(E_3G/At_1G) = $\frac{P(E_{3G}\cap At_{1G})}{P(At_{1G})}$ = $\frac{\frac{1}{5}}{\frac{4}{5}}$ = $\frac{1}{4}$

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(y)...

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Oh.. Conditional probability :/

Thanks :)

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@akash

how 4/15?

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@srestha, I was doing this way!!!

in question, there are 4 children

G G G G

G G G B

......

B B B B

So total 16 possible combinations out of this one thing is already given that at least one of them is a girl, that's why I have removed one combination (B B B B).Now there are 15 combinations are remaining and probability will be equally distributed to all 15 combinations.

And the asked probability is exactly 3 girls are required, so such cases would be C(4,3).

So required probability = C(4,3) /15 = 4/15, in this way i was getting..

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but is GBBB, BBBG is anytning different?

why r u taking combination?

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yes both things are same but I was considering place values of children but in this question, it doesn't require right??

Answer 1

all possibilities;-

(G,B) = (0,4), (1,3), (2,2), (3,1), (4,0)

it is given that :-

 atleast 1 out of 4 children is a girl:-

 now new sample space would be:-  (1,3), (2,2), (3,1), (4,0)

propability for exactly three girls:- means propability of (3,1):-  i.e  1/4

Comments

@Manu @hs_yadav

ans 4/15

Answer 2

Case where there is 1 girl-  4

Case when there are 2 girls- 6

Case when there are 3 girls- 4

Case when there are 4 girls- 1

total cases= 15

P(with exactly 3 girls)= 4/15