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Probability

by avenus » Tue May 05, 2009 3:00 pm
have a go at this. I don't quite agree with OE


Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl).

(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11
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Re: Probability

by dtweah » Tue May 05, 2009 4:24 pm
avenus wrote:have a go at this. I don't quite agree with OE


Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl).

(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11
Let the probability of having a girl be Head and that of having a boy be Tail. Then we are to find the probability of landing two heads in four tosses of a coin. Four tosses is the same as four attempts at birth, with head or tail, or boy or girl occuring with prob 1/2 each. But we are told that she has at least 2 girls. So you are finding the probability she has two boys Given she has at least two Girls, a reduced sample space or a conditional probability.

Let B= event of having two boys
G= Event of having at least 2 girls.

P(B|G)=P(B intercept G)/P(G).

Imagine the HHHT kind of arrangement replacing H with B and T with G. YOu know there are 16 such arrangement in the sample space. P(B int G) will be 6 of those 16. ggbb gbgb gbbg bggb bgbg bbgg. and P(G) will be 11 events out of 16: gggg gggb ggbg ggbb gbgg gbgb gbbg bggg bggb bgbg bbgg .
Putting all together we have P(B|G)= 6/16/11/16=6/11.

The trap is 4C2/2^4=3/8. This treatment is not conditional on at least 2 girls. The 5/11 is just in case you miscount.
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by preetosh » Tue May 05, 2009 9:33 pm
To me it seems 1/4 .
As two spots are fixed for girls and on two places , the probability of two boys would be 1/4

Let me know the answer.
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Re: Probability

by Vemuri » Tue May 05, 2009 10:10 pm
Based on the information provided, the following are the possibilities:

G G G G --> can be arranged in 1 way
G G G B --> can be arranged in 4 ways
G G B B --> can be arranged in 6 way6

So, the probability to have 2 boys = 6/(6+4+1) = 6/11 (E)
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by criszerriny » Wed May 06, 2009 4:49 am
Vamuri, can you present and explain a formula you used for how you were able to derive how the arrangements were calculated? For example, GGGG-arranged in 1 way, i understand the logic, but is there a formula?

Thanks
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by criszerriny » Wed May 06, 2009 4:51 am
Ahhh, nvm, I got it.

GGBB-Arranged in 6 ways.... 4C2= 4*3/2*1

Thank you
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by avenus » Wed May 06, 2009 6:49 am
Let the probability of having a girl be Head and that of having a boy be Tail. Then we are to find the probability of landing two heads in four tosses of a coin. Four tosses is the same as four attempts at birth, with head or tail, or boy or girl occuring with prob 1/2 each. But we are told that she has at least 2 girls. So you are finding the probability she has two boys Given she has at least two Girls, a reduced sample space or a conditional probability.

Let B= event of having two boys
G= Event of having at least 2 girls.

P(B|G)=P(B intercept G)/P(G).

Imagine the HHHT kind of arrangement replacing H with B and T with G. YOu know there are 16 such arrangement in the sample space. P(B int G) will be 6 of those 16. ggbb gbgb gbbg bggb bgbg bbgg. and P(G) will be 11 events out of 16: gggg gggb ggbg ggbb gbgg gbgb gbbg bggg bggb bgbg bbgg .
Putting all together we have P(B|G)= 6/16/11/16=6/11.

The trap is 4C2/2^4=3/8. This treatment is not conditional on at least 2 girls. The 5/11 is just in case you miscount
pretty much what I was looking for. Great explanation. Thanks.

OA is E
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