Chapter 10.1, Problem 50E

Solution

(a.)

The probability that a sophomore does not take either course.

It has been determined that the probability that a sophomore does not take either course is $10\%$ .

Given:

At a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a sophomore taking Spanish and let $B$ denote the event of a sophomore taking biology.

It is given that at a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Then, $P\left(A\right)=75\%$ , $P\left(B\right)=60\%$ and $P\left(A\cap B\right)=45\%$ .

Now, the probability that a sophomore does not take either course is $P\left({\left(A\cup B\right)}^C\right)$ .

Now, put $P\left(A\right)=75\%$ , $P\left(B\right)=60\%$ and $P\left(A\cap B\right)=45\%$ in $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ to get,

  $\begin{array}{c}P\left(A\cup B\right)+45\%=75\%+60\%\\ P\left(A\cup B\right)+45\%=135\%\\ P\left(A\cup B\right)=135\%-45\%\\ P\left(A\cup B\right)=90\%\end{array}$

Now, $P\left(A^C\right)=100\%-P\left(A\right)$ .

Then, $P\left({\left(A\cup B\right)}^C\right)=100\%-P\left(A\cup B\right)$ .

Put $P\left(A\cup B\right)=90\%$ in the above equation to get,

  $P\left({\left(A\cup B\right)}^C\right)=10\%$

So, the probability that a sophomore does not take either course is $10\%$ .

Conclusion:

It has been determined that the probability that a sophomore does not take either course is $10\%$ .

(b.)

The probability that a sophomore takes Spanish or biology.

It has been determined that the probability that a sophomore takes Spanish or biology is $45\%$ .

Given:

At a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a sophomore taking Spanish and let $B$ denote the event of a sophomore taking biology.

It is given that at a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Then, $P\left(A\right)=75\%$ , $P\left(B\right)=60\%$ and $P\left(A\cap B\right)=45\%$ .

Now, the probability that a sophomore takes Spanish or biology is $P\left(A\cup B\right)-P\left(A\cap B\right)$ .

As determined previously, $P\left(A\cup B\right)=90\%$ .

Put $P\left(A\cup B\right)=90\%$ and $P\left(A\cap B\right)=45\%$ in $P\left(A\cup B\right)-P\left(A\cap B\right)$ to get,

  $P\left(A\cup B\right)-P\left(A\cap B\right)=90\%-45\%$

Simplifying,

  $P\left(A\cup B\right)-P\left(A\cap B\right)=45\%$

So, the probability that a sophomore takes Spanish or biology is $45\%$ .

Conclusion:

It has been determined that the probability that a sophomore takes Spanish or biology is $45\%$ .

(c.)

The probability that a sophomore takes biology given that she takes Spanish.

It has been determined that the probability that a sophomore takes biology given that she takes Spanish $60\%$ .

Given:

At a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a sophomore taking Spanish and let $B$ denote the event of a sophomore taking biology.

It is given that at a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Then, $P\left(A\right)=75\%$ , $P\left(B\right)=60\%$ and $P\left(A\cap B\right)=45\%$ .

The probability that a sophomore takes biology given that she takes Spanish is $\frac{P\left(A\cap B\right)}{P\left(A\right)}$ .

Put $P\left(A\cap B\right)=45\%$ and $P\left(A\right)=75\%$ in $\frac{P\left(A\cap B\right)}{P\left(A\right)}$ to get,

  $\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{45\%}{75\%}$

Simplifying,

  $\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{3}{5}$

Converting to percentage,

  $\frac{P\left(A\cap B\right)}{P\left(A\right)}=60\%$

So, the probability that a sophomore takes biology given that she takes Spanish $60\%$ .

Conclusion:

It has been determined that the probability that a sophomore takes biology given that she takes Spanish $60\%$ .

(d.)

The probability that a sophomore takes Spanish given that she takes biology.

It has been determined that the probability that a sophomore takes Spanish given that she takes biology is $75\%$ .

Given:

At a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a sophomore taking Spanish and let $B$ denote the event of a sophomore taking biology.

It is given that at a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Then, $P\left(A\right)=75\%$ , $P\left(B\right)=60\%$ and $P\left(A\cap B\right)=45\%$ .

The probability that a sophomore takes Spanish given that she takes biology is $\frac{P\left(A\cap B\right)}{P\left(B\right)}$ .

Put $P\left(A\cap B\right)=45\%$ and $P\left(B\right)=60\%$ in $\frac{P\left(A\cap B\right)}{P\left(B\right)}$ to get,

  $\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{45\%}{60\%}$

Simplifying,

  $\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{3}{4}$

Converting to percentage,

  $\frac{P\left(A\cap B\right)}{P\left(B\right)}=75\%$

So, the probability that a sophomore takes Spanish given that she takes biology is $75\%$ .

Conclusion:

It has been determined that the probability that a sophomore takes Spanish given that she takes biology is $75\%$ .

(e.)

If among sophomores, taking Spanish and taking biology are independent events.

It has been determined that among sophomores, taking Spanish and taking biology are independent events.

Given:

At a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a sophomore taking Spanish and let $B$ denote the event of a sophomore taking biology.

It is given that at a large high school, $75\%$ of the sophomores take Spanish, $60\%$ take biology, and $45\%$ take both courses.

Then, $P\left(A\right)=75\%$ , $P\left(B\right)=60\%$ and $P\left(A\cap B\right)=45\%$ .

Now, if $A$ and $B$ are independent events, then $P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$ .

Put $P\left(A\right)=75\%$ and $P\left(B\right)=60\%$ in $P\left(A\right)P\left(B\right)$ to get,

  $\begin{array}{c}P\left(A\right)P\left(B\right)=\left(75\%\right)\left(60\%\right)\\ =\left(\frac{75}{100}\right)\left(\frac{60}{100}\right)\\ =\left(\frac{75}{100}\right)\left(\frac{60}{100}\right)\times 100\%\\ =\left(\frac{75}{100}\right)\left(60\right)\%\end{array}$

Solving,

  $P\left(A\right)P\left(B\right)=45\%$

It is given that $P\left(A\cap B\right)=45\%$ .

Therefore, clearly $P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$ .

This implies that $A$ and $B$ are independent events.

Thus, among sophomores, taking Spanish and taking biology are independent events.

Conclusion:

It has been determined that among sophomores, taking Spanish and taking biology are independent events.