Get the answer to "Let the random variable X denote the number..."

dohtarjev510

Answered question

2022-02-14

Let the random variable X denote the number of girls in a five-child family. If the probability of a female birth is 0.5, what are the probabilities of having 0, 1, 2, 3, 4, and 5 girls?

Answer & Explanation

Bordadwayjpf

Beginner2022-02-15Added 17 answers

Step 1
The probability of having a girl is given as 0.5. This means that the probability of not having a girl is also 0.5.
The last thing to look at is the number of ways we can achieve any given result. For instance, there is only 1 way to have 0 girls (and the same goes for 5 girls), but there are 5 different ways we can have 1 girl (and the same goes for 4 girls). This is a combinations question, with:

C_{n, k} = (\begin{matrix} n \\ k \end{matrix}) = \frac{n!}{(k!) (n - k)!}

with

n = population, k = picks

Boiling it all down, we end up with:
0 girls

= (\begin{matrix} 5 \\ 0 \end{matrix}) {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{5} = 1 \times 1 \times \frac{1}{32} = \frac{1}{32}

1 girls

= (\begin{matrix} 5 \\ 1 \end{matrix}) {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{4} = 5 \times \frac{1}{2} \times \frac{1}{16} = \frac{5}{32}

2 girls

= (\begin{matrix} 5 \\ 2 \end{matrix}) {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{3} = 10 \times \frac{1}{4} \times \frac{1}{8} = \frac{10}{32} = \frac{5}{16}

3 girls

= (\begin{matrix} 5 \\ 3 \end{matrix}) {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{2} = 10 \times \frac{1}{8} \times \frac{1}{4} = \frac{10}{32} = \frac{5}{16}

4 girls

= (\begin{matrix} 5 \\ 4 \end{matrix}) {(\frac{1}{2})}^{4} {(\frac{1}{2})}^{1} = 5 \times \frac{1}{16} \times \frac{1}{2} = \frac{5}{32}

5 girls

= (\begin{matrix} 5 \\ 5 \end{matrix}) {(\frac{1}{2})}^{5} {(\frac{1}{2})}^{0} = 1 \times \frac{1}{32} \times 1 = \frac{1}{32}

Notice that if we add all of these up, we get 1 (or the sum total of all probabilities):

\frac{1}{32} + \frac{5}{32} + \frac{10}{32} + \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{32}{32} = 1

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