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
Probability--1 boy & two girls
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total combo/combo of girls=3!/2!=3prachich1987 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
P(BGG)=(1/2)^3=1/8
1/8 * 3=3/8
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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".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
In other words, the question is identical to:
Once you recognize it as a coin flip problem, you can apply the coin flip formula to solve:If a fair coin is flipped 3 times, what's the probability of getting exactly two heads?
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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- prachich1987
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@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.
Well,my question may sound stupid.
But, I just wanted to confirm that here we need not consider Girl1 & Girl2 different.
Thanks!
Prachi
Prachi
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P(B)=P(G)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(G)=P(G)
doesn't matter actually here; otherwise we had to go all way around - or seek complement probabilities.
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Hi,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.
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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We need to determine the probability for a family with three children to have a boy and two girls.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
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
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