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Q28E

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Discrete Mathematics and its Applications
Found in: Page 467
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

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

  1. Exactly three boys?
  2. At least one boy?
  3. At least one girl?
  4. 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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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 C(n,k)pkqn-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-pq=1-0.51q=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(n,k)pkqn-k=C(5,3)p3q5-3C(5,3)p3q5-3=C(5,3)(0.51)3(0.49)2C(5,3)p3q5-3=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(n,k)pkqn-k=C(5,5)p5q5-5C(5,5)p5q5-5=C(5,5)(0.49)5(0.51)0C(5,5)p5q5-5=(0.49)5

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(n,k)pkqn-k=C(5,5)p5q5-5C(5,5)p5q5-5=C(5,5)(0.49)0(0.51)5C(5,5)p5q5-5=(0.51)5

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(5,5)p5q5-5+C(5,5)p5q5-5=C(5,5)(0.49)0(0.51)5+C(5,5)(0.49)5(0.51)0C(5,5)p5q5-5+C(5,5)p5q5-5=C(5,5)(0.51)5+C(5,5)(0.49)5C(5,5)(0.51)5+C(5,5)(0.49)5=0.0627

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

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