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Q28E

Expert-verifiedFound in: Page 467

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Question: Assume that the probability a child is a boy is 0.51 and that the sexes of the children born into a family are independent. What is the probability that a family of five children has **

**Exactly three boys?****At least one boy?****At least one girl?****All children of the same sex?**

**Answer**

The probability of exactly three boys is 0.3184

The probability of at least one boy is =0.9717

The probability of at least one boy is =0.9654

The probability of all the children are of the same sex =0.9654

The probability of a child being boy is 0.51

**Definition: The probability of exactly k successes in n independent Bernoulli trials, with probability of success p and probability of failure q = 1-p, is ${\mathit{C}}{\mathbf{(}}{\mathit{n}}{\mathbf{,}}{\mathit{k}}{\mathbf{)}}{{\mathit{p}}}^{{\mathbf{k}}}{{\mathit{q}}}^{\mathbf{n}\mathbf{-}\mathbf{k}}$**

Given that the probability of child being boy is, p= 0.51

The probability of child being girl is, q = 1-p

$q=1-p\phantom{\rule{0ex}{0ex}}q=1-0.51\phantom{\rule{0ex}{0ex}}q=0.49$

Given that out of 5 children exactly three boys,

Hence using the Bernoulli trials, we get,

$C(n,k){p}^{k}{q}^{n-k}=C(5,3){p}^{3}{q}^{5-3}\phantom{\rule{0ex}{0ex}}C(5,3){p}^{3}{q}^{5-3}=C(5,3)(0.51{)}^{3}(0.49{)}^{2}\phantom{\rule{0ex}{0ex}}C(5,3){p}^{3}{q}^{5-3}=0.3184$

Therefore, the probability of exactly three boys is 0.3184

Given that out of 5 children at least one is a boy, so there can be 1, 2 ,3, 4, or 5 boys

Hence using the Bernoulli trials, we get,

$C(n,k){p}^{k}{q}^{n-k}=C(5,5){p}^{5}{q}^{5-5}\phantom{\rule{0ex}{0ex}}C(5,5){p}^{5}{q}^{5-5}=C(5,5)(0.49{)}^{5}(0.51{)}^{0}\phantom{\rule{0ex}{0ex}}C(5,5){p}^{5}{q}^{5-5}=(0.49{)}^{5}\phantom{\rule{0ex}{0ex}}$

Therefore, the probability of at least one boy is =0.9717

Given that out of 5 children at least one is girl, so there can be 1, 2 ,3, 4, or 5 girls

Hence using the Bernoulli trials, we get,

$C(n,k){p}^{k}{q}^{n-k}=C(5,5){p}^{5}{q}^{5-5}\phantom{\rule{0ex}{0ex}}C(5,5){p}^{5}{q}^{5-5}=C(5,5)(0.49{)}^{0}(0.51{)}^{5}\phantom{\rule{0ex}{0ex}}C(5,5){p}^{5}{q}^{5-5}=(0.51{)}^{5}\phantom{\rule{0ex}{0ex}}$

Therefore, the probability of at least one boy is =0.9654

Given that all the children are of the same sex

Hence using the Bernoulli trials, we get,

$C(5,5){p}^{5}{q}^{5-5}+C(5,5){p}^{5}{q}^{5-5}=C(5,5)(0.49{)}^{0}(0.51{)}^{5}+C(5,5)(0.49{)}^{5}(0.51{)}^{0}\phantom{\rule{0ex}{0ex}}C(5,5){p}^{5}{q}^{5-5}+C(5,5){p}^{5}{q}^{5-5}=C(5,5)(0.51{)}^{5}+C(5,5)(0.49{)}^{5}\phantom{\rule{0ex}{0ex}}C(5,5)(0.51{)}^{5}+C(5,5)(0.49{)}^{5}=0.0627\phantom{\rule{0ex}{0ex}}$

Therefore, the probability of all the children are of the same sex =0.9654

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