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Who eats breakfast?Students in an urban school were curious about how many children regularly eat breakfast. They conducted a survey, asking, 鈥淒o you eat breakfast on a regular basis?鈥 All 595students 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 Fas getting a female student and event Bas getting a student who eats breakfast regularly.

a. Find P(BC)

b. Find P(FandBC). Interpret this value in context.

c. Find P(ForBC).

Short Answer

Expert verified

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 is0.6807

Step by step solution

01

Part (a) Step 1 : Given Information

We have to compute the probability of BC

02

Part (a) Step 2 : Simplification

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

Fis the occurrence of a female student being chosen.
Bis the occurrence of a student who frequently eats breakfast.
The following formula was used:

P(BC)=NumberofstudentswhodonoteatbreakfastregularlyTotalnumberofstudents

BCrefers to a student who does not eat breakfast on a regular basis.

Between the two tables,
295kids do not eat breakfast on a regular basis.
There are 595pupils in all.

P(Bc)=295595=0.4958

03

Part (b) Step 1 : Given Information

We have to compute the probability of Fand Bc

04

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(FandBc)=165595=0.2773

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

05

Part (c) Step 1 : Given Information

We have to compute P(FandBc).

06

Part (c) Step 2 : Simplification

The formula we use :

P(ForBc)=P(F)+P(Bc)P(FandBc)P(F)=NumberoffemalestudentsTotalnumberofstudents

There are 275female students.
There are 595pupils in all.

P(F)=275595=0.4622P(FandBc)=0.4622+0.49580.2773=0.6807

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