Question

Question: Let E and F be the events that a family of$\mathit{n}$children has children of both sexes and has at most one boy, respectively. Are E and F Independent if

role="math" localid="1668598698436" $$\begin{gathered} \mathit{a}\mathbf{)}\mathbf{}\mathit{n}\mathbf{=}\mathbf{2} \\ \mathit{b}\mathbf{)}\mathit{n}\mathbf{=}\mathbf{4}\mathbf{?} \\ \mathit{c}\mathbf{)}\mathbf{=}\mathbf{5}\mathbf{?} \end{gathered}$$

Short answer

Answer

E and F are not Independent Events.

E and F are not Independent Events.

E and F are not Independent Events.

Step-by-step solution

Given Information

1. E and F be the events that a family of n=2 children of both sexes and has at most one boy.
2. E and F be the events that a family of n=4 children of both sexes and has at most one boy.
3. E and F be the events that a family of n=5 children of both sexes and has at most one boy.

Definition of Probability and Independent events

In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event does affect the probability of the other event, then the events are dependent.

Calculating the Probability

Here, E is the event that the family of ‘2’ children has children of both sexes.

When a child is born it can be a male child or a female child. The chances of each child are equally likely.

$E=\left\{MF,FM\right\}$

Let S be the set of all possible outcomes of the two children.

$$\begin{gathered} S=\left\{MM,FF,MF,FM\right\} \\ \left|E\right|=2and\left|S\right|=4 \end{gathered}$$

Hence, the probability that the family has both children is

$P\left(E\right)=\frac{\left|E\right|}{\left|S\right|}=\frac{3}{4}------\left(1\right)$

Given ‘F’ is the event that the family has at most one boy i.e.

$F=\left\{MF,FM,FF\right\}$

The Probability that the family has at most one boy is

$P\left(F\right)=\frac{\left|F\right|}{\left|S\right|}=\frac{3}{4}------\left(2\right)$

Now, we have to find out E and F are Independent events.

We have to first find $E\cap F$

i.e.,

$$\begin{gathered} E\cap F=\left\{MF,FM\right\} \\ \left|E\cap F\right|=2 \\ P\left(E\cap F\right)=\frac{\left|E\cap F\right|}{\left|S\right|} \\ =\frac{2}{4}=\frac{1}{2} \\ Now,P\left(E\right)\times P\left(F\right)=\frac{1}{2}\times \frac{3}{4}=\frac{3}{8}u\sin g\left(1\right)and\left(2\right) \\ AsP\left(E\cap F\right)\ne P\left(E\right)\times P\left(F\right) \end{gathered}$$

Hence, E and F are not Independent events.

Calculating the Probability

Here, E is the event that the family of ‘4’ children has children of both sexes.

$$\begin{gathered} P\left(E\right)=1-\left[\mathrm{Pr}obabilityofallmalechildren+\mathrm{Pr}obabilityofallfemalechildren\right] \\ =1-\left[C\left(4,4\right){\left(\frac{1}{2}\right)}^4+C\left(4,4\right){\left(\frac{1}{2}\right)}^4\right] \\ =1-\left[\frac{1}{8}\right]=\frac{7}{8} \end{gathered}$$

Let F be the event that a family of 4 children has at most one boy.

$$\begin{gathered} P\left(E\right)=1-\left[\mathrm{Pr}obabilityofoneboyand{2}^{3}girls+\mathrm{Pr}obabilityofnoboy\right] \\ {=}^4{C}_1{\left(\frac{1}{2}\right)}^4{+}^4{C}_4{\left(\frac{1}{2}\right)}^4 \\ =4\times \left(\frac{1}{16}\right)+\frac{1}{16} \\ =\frac{5}{16} \end{gathered}$$

Now, we have to find out E and F are Independent events

We have to first find

i.e.,

$$\begin{gathered} E\cap F=\left\{BGGG,GBGG,GGBG,GGGB\right\} \\ P\left(E\cap F\right)=4\times {\left(\frac{1}{2}\right)}^2 \\ =\frac{4}{16}=\frac{1}{4}------\left(1\right) \end{gathered}$$

Now, we find

$$\begin{gathered} P\left(E\right)\times P\left(F\right)=\frac{7}{8}\times \frac{5}{6} \\ =\frac{35}{128}-----\left(2\right) \\ Nowfrom\left(1\right)and\left(2\right)weconclu\det hat, \\ P\left(E\cap F\right)\ne P\left(E\right)\times P\left(F\right) \end{gathered}$$

Hence, E and F are not Independent events

Calculating the Probability

Here, E is the event that the family of ‘5’ children has children of both sexes.

$$\begin{gathered} P\left(E\right)=1-\left[\mathrm{Pr}obabilityofallmalechildren+\mathrm{Pr}obabilityofallfemalechildren\right] \\ =1-\left[5C_5{\left(\frac{1}{2}\right)}^5+5C_5{\left(\frac{1}{2}\right)}^5\right] \\ =1-\left[\frac{1}{32}+\frac{1}{32}\right] \\ =1-\left(\frac{2}{32}\right) \\ =1-\frac{1}{16} \\ =\frac{15}{16} \end{gathered}$$

Let F be the event that a family of 5 children has at most one boy.

$$\begin{gathered} F=\left\{BGGGG,GBGGG,GGBGG,GGGBG,GGGGB,GGGGG\right\} \\ =C\left(5,1\right){\left(\frac{1}{2}\right)}^5+{\left(\frac{1}{2}\right)}^5 \\ =\left(\frac{6}{32}\right)=\frac{3}{16} \end{gathered}$$

Now, we have to find out E and F are Independent events

We have to first find $E\cap F$

i.e.,

$$\begin{gathered} E\cap F=\left\{BGGGG,GBGGG,GGGBG,GGGGB,GGBGG\right\} \\ \left|E\cap F\right|=5 \\ P\left(E\cap F\right)=\frac{5}{32} \\ Nowputtingvalueof\left(1\right)and\left(2\right) \\ P\left(E\right)\times P\left(F\right)=\frac{15}{16}\times \frac{3}{16} \\ =\frac{45}{{16}^2} \end{gathered}$$

$AsP\left(E\cap F\right)\ne P\left(E\right)\times P\left(F\right)$

Hence, E and F are not Independent events.