A couple wants four children. If the probability that, on each birthing, the baby is boy or girl is equal to each other, what is the probability the they will exactly produce two boys and two girls?
oa [spoiler]3/8[/spoiler]
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GMATGuruNY
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P(exactly n times) = P(one way) * total possible ways.mariah wrote:A couple wants four children. If the probability that, on each birthing, the baby is boy or girl is equal to each other, what is the probability the they will exactly produce two boys and two girls?
oa [spoiler]3/8[/spoiler]
P(one way):
One way to get an equal number of boys and girls is for first two children born to be boys and for the last two children born to be girls.
P(BBGG) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16.
Total possible ways:
Any arrangement of the letters BBGG will yield exactly 2 boys and 2 girls.
Thus, to account for all the ways to get exactly 2 boys and 2 girls, the result above needs to be multiplied by the number of ways to arrange BBGG.
Number of ways to arrange BBGG = 4!/(2!2!) = 6.
Multiplying the results above, we get:
P(exactly 2 boys and 2 girls) = 6 * 1/16 = 3/8.
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ronnie1985
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Binomial distribution.
p = 1/2 p = 1/2
n = 4 r =2
P = nCrp^rq^(n-r) = 4C2 (1/2)^2 * (1/2)^2 = 3/8
p = 1/2 p = 1/2
n = 4 r =2
P = nCrp^rq^(n-r) = 4C2 (1/2)^2 * (1/2)^2 = 3/8
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