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Expert-verified Found in: Page 467 ### Discrete Mathematics and its Applications

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?

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

See the step by step solution

## Step 1: Given data

The probability of a child being boy is 0.51

## Step 2: Definitions and formula to be used

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 ${\mathbit{C}}{\mathbf{\left(}}{\mathbit{n}}{\mathbf{,}}{\mathbit{k}}{\mathbf{\right)}}{{\mathbit{p}}}^{{\mathbf{k}}}{{\mathbit{q}}}^{\mathbf{n}\mathbf{-}\mathbf{k}}$

## Step 3: To find probability of child being girl

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

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

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

## Step 4: Evaluation of probability of exactly three boys

Given that out of 5 children exactly three boys,

Hence using the Bernoulli trials, we get,

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

Therefore, the probability of exactly three boys is 0.3184

## Step 5: Evaluation of probability of at least one boy

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,

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

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

## Step 6: Evaluation of probability of at least one girl

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,

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

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

## Step 7: Evaluation of probability of all children are of same sex

Given that all the children are of the same sex

Hence using the Bernoulli trials, we get,

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

Therefore, the probability of all the children are of the same sex =0.9654 ### Want to see more solutions like these? 