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Probability--1 boy & two girls

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by prachich1987 » Tue Feb 01, 2011 5:30 am
What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8
b) ¼
c) ½
d) 3/8
e) 5/8

Source : some online material
Thanks!
Prachi
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by maihuna » Tue Feb 01, 2011 5:34 am
I think it will be total arrangeent of XYY out of total of 8 arrangements

See : XYY, YYX, YXY

so 3/8
Charged up again to beat the beast :)
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by Night reader » Tue Feb 01, 2011 9:04 am
prachich1987 wrote:What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8
b) ¼
c) ½
d) 3/8
e) 5/8

Source : some online material
total combo/combo of girls=3!/2!=3
P(BGG)=(1/2)^3=1/8
1/8 * 3=3/8
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by Stuart@KaplanGMAT » Tue Feb 01, 2011 1:21 pm
prachich1987 wrote:What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8
b) ¼
c) ½
d) 3/8
e) 5/8

Source : some online material
The explanations provided are great, I just wanted to add that this is an example of what we sometimes call a "pseudo-coin flip question".

In other words, the question is identical to:
If a fair coin is flipped 3 times, what's the probability of getting exactly two heads?
Once you recognize it as a coin flip problem, you can apply the coin flip formula to solve:

The probability of getting k results on n flips is:

nCk/2^n

(nCk = n!/k!(n-k)!, the combinations formula).

For this particular question, the other solutions may very well be quicker (remember, just because there is a fancy formula doesn't mean that algebra is the best approach to a problem), but the formula is useful on more advanced coin flip/pseudo-coin flip questions.
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by prachich1987 » Wed Feb 02, 2011 8:16 pm
@Stuart Kovinsky : Thanks..I liked this pseudo coin flip approach.
Well,my question may sound stupid.
But, I just wanted to confirm that here we need not consider Girl1 & Girl2 different.
Thanks!
Prachi
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by Night reader » Wed Feb 02, 2011 8:30 pm
prachich1987 wrote:@Stuart Kovinsky : Thanks..I liked this pseudo coin flip approach.
Well,my question may sound stupid.
But, I just wanted to confirm that here we need not consider Girl1 & Girl2 different.
P(B)=P(G)
P(G)=P(G)

doesn't matter actually here; otherwise we had to go all way around - or seek complement probabilities.
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by Stuart@KaplanGMAT » Thu Feb 03, 2011 8:10 am
prachich1987 wrote:@Stuart Kovinsky : Thanks..I liked this pseudo coin flip approach.
Well,my question may sound stupid.
But, I just wanted to confirm that here we need not consider Girl1 & Girl2 different.
Hi,

we're not arranging specific boys and specific girls, we're just choosing 1 boy out of 3 children. Since we don't care about the specific children, we can use the simple combinations formula.
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hi

by Jeff@TargetTestPrep » Mon Dec 18, 2017 11:40 am
prachich1987 wrote:What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8
b) ¼
c) ½
d) 3/8
e) 5/8
We need to determine the probability for a family with three children to have a boy and two girls.

P(G-G-B) = 1/2 x 1/2 x 1/2 = 1/8

We also see that there are 3 ways to arrange two girls and one boy:

G-G-B

G-B-G

B-G-G

Thus, the probability is 1/8 x 3 = 3/8.

Answer: D

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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