Question

Who eats breakfast?Students in an urban school were curious about how many children regularly eat breakfast. They conducted a survey, asking, “Do you eat breakfast on a regular basis?” All $595$students in the school responded to the survey. The resulting data are shown in the two-way table.

Suppose we select a student from the school at random. Define event $F$as getting a female student and event $B$as getting a student who eats breakfast regularly.

a. Find $P\left(B^C\right)$

b. Find $P(Fand{B}^C)$. Interpret this value in context.

c. Find $P(For{B}^C)$.

Short answer

a. Student who does not eat breakfast on a regular basis $0.4958$

b. The probability that a student is female and do not eats breakfast regularly is $0.2773$

c. The probability that a student is female or do not eats breakfast regularly is$0.6807$

Step-by-step solution

Part (a) Step 1 : Given Information

We have to compute the probability of $B^C$

Part (a) Step 2 : Simplification

Below is a two-way table showing gender and breakfast habits.

$F$is the occurrence of a female student being chosen.

$B$is the occurrence of a student who frequently eats breakfast.

The following formula was used:

$\hspace{1em}P\left(B^C\right)=\frac{Numberofstudentswhodonoteatbreakfastregularly}{Totalnumberofstudents}$

$B^C$refers to a student who does not eat breakfast on a regular basis.

Between the two tables,

$295$kids do not eat breakfast on a regular basis.

There are $595$pupils in all.

$$\begin{gathered} P\left(B^c\right)=\frac{295}{595} \\ =0.4958 \end{gathered}$$

Part (b) Step 1 : Given Information

We have to compute the probability of $F$and $B^c$

Part (b) Step 2 : Simplification

Female students do not eat breakfast on a daily basis is $165$

The total number of pupils in the class is $595$

$P(Fand{B}^c)=\frac{165}{595}=0.2773$

The probability that a student is female and do not eats breakfast regularly is $0.2773$.

Part (c) Step 1 : Given Information

We have to compute $P(Fand{B}^c)$.

Part (c) Step 2 : Simplification

The formula we use :

$$\begin{gathered} P(For{B}^c)=P\left(F\right)+P\left(B^c\right)-P(Fand{B}^c) \\ \hspace{1em}P\left(F\right)=\frac{Numberoffemalestudents}{Totalnumberofstudents} \end{gathered}$$

There are $275$female students.

There are $595$pupils in all.

$$\begin{gathered} P\left(F\right)=\frac{275}{595}=0.4622 \\ P(Fand{B}^c)=0.4622+0.4958-0.2773 \\ =0.6807 \end{gathered}$$