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Chapter 3: Problem 73

Suppose that each child born to a couple is equally likely to be a boy or a girl, independently of the sex distribution of the other children in the family. For a couple having 5 children, compute the probabilities of the following events: (a) All children are of the same sex. (b) The 3 eldest are boys and the others girls. (c) Exactly 3 are boys. (d) The 2 oldest are girls. (e) There is at least 1 girl.

Short Answer

Expert verified
(a) The probability that all the children are of the same sex is 0.0625. (b) The probability that the three eldest are boys and the others are girls is 0.03125. (c) The probability that exactly three of the children are boys is 0.3125. (d) The probability that the two oldest are girls is 0.25. (e) The probability that there is at least one girl among the five children is 0.96875.

Step by step solution

01

Event a: All children are of the same sex

There are two cases for this event: all children are boys, or all children are girls. We will find the individual probabilities and then add them up since these are mutually exclusive events. 1. All children are boys: \(P(X=5) = \binom{5}{5}(0.5)^5(0.5)^{5-5} = 0.03125\) 2. All children are girls: \(P(X=0) = \binom{5}{0}(0.5)^0(0.5)^{5-0} = 0.03125\) So, the probability of all children being of the same sex is the sum of the probabilities of all boys and all girls: \(P(A) = 0.03125 + 0.03125 = 0.0625\)
02

Event b: The 3 eldest are boys and the others girls

This is a specific event where the first 3 children are boys and the last 2 are girls. Therefore, we can directly multiply the probabilities for each child: \(P(B) = (0.5)^3 \times (0.5)^2 = 0.5^5 = 0.03125\)
03

Event c: Exactly 3 children are boys

For this event, we will use the binomial probability formula with k=3: \(P(C) = \binom{5}{3}(0.5)^3(0.5)^{5-3} = 10 \times 0.125 \times 0.25 = 0.3125\)
04

Event d: The 2 oldest are girls

In this case, we don't care about the genders of the last 3 children, so we only need to find the probability of having 2 consecutive girls: \(P(D) = 0.5^2 = 0.25\)
05

Event e: There is at least 1 girl

Instead of finding probabilities for each case with at least 1 girl, we will find the probability of having all boys and then subtract it from 1. Probability of having all boys: \(P(E) = 1 - P(\text{all boys}) = 1 - 0.03125 = 0.96875\) So, the probabilities for each event are: a) All children are of the same sex: 0.0625 b) The 3 eldest are boys and the others girls: 0.03125 c) Exactly 3 are boys: 0.3125 d) The 2 oldest are girls: 0.25 e) There is at least 1 girl: 0.96875

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