Approximately $281,000,000$people over age five live in the United States. Of these people, $55,000,000$speak a language other than English at home. Of those who speak another language at home, $62.3\%$speak Spanish.

Let:E = speaks English at home; E′ = speaks another language at home; S = speaks Spanish;

Finish each probability statement by matching the correct answer.

Approximately $281000000$people over age five live in the United States. Of these people, $55000000$speak a language other than English at home. Of those who speak another language at home, $62.3\%$speak Spanish.

Consider the below events:

$\text{E}=$speaks English at home

$\text{E}\text{'}=$speaks another language at home

$\text{S}=$speaks English

Also, we have

Total number of people$=281000000$

Number of people who speak a language other than English at home$=55000000$

$P\left(S|E\text{'}\right)=62.3\%$

First, we need to calculate $P\left(E\text{'}\right)$localid="1648128504606" $$\begin{gathered} P\left(E^{\text{'}}\right)=\frac{\text{Number of people whospeak a language other than English at home}}{\text{Total number of people}} \\ P\left(E^{\text{'}}\right)=\frac{55000000}{281000000} \\ P\left(E\text{'}\right)=0.1957 \end{gathered}$$

Now, calculating $P\left(E\right)$

$$\begin{gathered} P\left(E\right)=1-P\left(E\text{'}\right) \\ P\left(E\right)=1-0.1957 \\ P\left(E\right)=0.8043 \end{gathered}$$

Again, calculating $P\left(SandE\text{'}\right)$

$$\begin{gathered} P\left(SAND{E}^{\text{'}}\right)=P\left(S\mid E^{\text{'}}\right)\times P\left(E^{\text{'}}\right) \\ P\left(SAND{E}^{\text{'}}\right)=0.623\times 0.1957 \\ P\left(SAND{E}^{\text{'}}\right)=0.1219 \end{gathered}$$

Last, we need to find $P\left(S|E\text{'}\right)$

Given that $P\left(S|E\text{'}\right)=62.3\%$

$P\left(S|E\text{'}\right)=0.623$