Question

(a) Acandy vending machine is out of order. The probability that you get a candybar (with or without return of your money) is$\frac{\mathbf{1}}{\mathbf{2}}$, the probability that you getyour money back (with or without candy) is $\frac{\mathbf{1}}{\mathbf{2}}$, and the probability that youget both the candy and your money back is $\frac{\mathbf{1}}{\mathbf{12}}$. What is the probability that youget nothing at all? Suggestion: Sketch a geometric diagram similar to Figure 3.1, indicate regions representing the various possibilities and their probabilities; then set up a four-point sample space and the associated probabilities of the points.

(b) Suppose you try again to get a candy bar as in part (a). Set up the 16-point

sample space corresponding to the possible results of your two attempts tobuy a candy bar, and find the probability that you get two candy bars (andno money back); that you get no candy and lose your money both times; thatyou just get your money back both times.

Short answer

Answer

(a) The probability of getting nothing is $\frac{1}{4}$.

(b) The probability to get two candy bars without getting any money back is role="math" localid="1654777977384" $P\left(AAB\text{'}B\text{'}\right)=\frac{25}{144};$probability to not get candy and also lose your money both times is role="math" localid="1654777997726" $P\left(A\text{'}A\text{'}B\text{'}B\text{'}\right)=\frac{1}{16}$; that to just get your money back both times is $P\left(A\text{'}A\text{'}B\text{'}B\text{'}\right)=\frac{1}{16}$.

Step-by-step solution

Given Information

The probability that you get a candy bar (with or without return of your money) is $\frac{1}{2}$, the probability that you get your money back (with or without candy) is $\frac{1}{3}$, and the probability that you get both the candy and your money back is $\frac{1}{12}$.

Definition of Independent Event

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

When events are independent, apply the formula $P\left(AB\right)=P\left(A\right)·P\left(B\right)$where A and B are the events.

Draw the region for the given scenario

Let A be the event of getting a bar and B be the event of getting the money back.

Thus $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3}$and $P\left(AB\right)=\frac{1}{12}$.

Draw the region depicting the respective probabilities.

Creating Sample Space for experiment when two die are rolled

Let A be the event of getting a bar and B be the event of getting the money back, thus A' and B' would mean not getting bar and money back respectively.

There are 4 points in the sample space.

Find the sample space for the given problem.

$\mathit{A}\mathit{B}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{A}\mathbf{\text{'}}\mathit{B}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{A}\mathit{B}\mathbf{\text{'}}\mathbf{}\mathbf{}\mathbf{}\mathit{A}\mathbf{\text{'}}\mathit{B}\mathbf{\text{'}}$

Finding the probability of getting nothing

The probability of getting nothing can be found by subtracting the probability of getting at least one thing from 1.

Find the probability of getting money or bar or both, that is $P\left(A+B\right)$, using the formula $P\left(A+B\right)=P\left(A\right)+P\left(B\right)-P\left(B\right)-\left(AB\right)$.

$$\begin{gathered} P\left(A+B\right)=\frac{1}{2}+\frac{1}{3}-\frac{1}{12} \\ =\frac{6+4-1}{12} \\ =\frac{3}{4} \end{gathered}$$

Find the probability of getting nothing

$$\begin{gathered} P\left(A\text{'}B\text{'}\right)=1-\frac{3}{4} \\ =\frac{1}{4} \end{gathered}$$

Creating the sample space when two attempts to buy a candy bar is made

There are 16 points in the sample space.

Find the sample space for the given problem.

$\begin{array}{cccc}\mathbf{AABB}& \mathbf{AABB}\mathbf{\text{'}}& \mathbf{AAB}\mathbf{\text{'}}\mathbf{B}& \mathbf{AAB}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\\ \mathbf{A}\mathbf{\text{'}}\mathbf{ABB}& \mathbf{A}\mathbf{\text{'}}\mathbf{ABB}\mathbf{\text{'}}& \mathbf{A}\mathbf{\text{'}}\mathbf{AB}\mathbf{\text{'}}\mathbf{B}& \mathbf{A}\mathbf{\text{'}}\mathbf{AB}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\\ \mathbf{AA}\mathbf{\text{'}}\mathbf{BB}& \mathbf{AA}\mathbf{\text{'}}\mathbf{BB}\mathbf{\text{'}}& \mathbf{AA}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\mathbf{B}& \mathbf{AA}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\\ \mathbf{A}\mathbf{\text{'}}\mathbf{A}\mathbf{\text{'}}\mathbf{BB}& \mathbf{A}\mathbf{\text{'}}\mathbf{A}\mathbf{\text{'}}\mathbf{BB}\mathbf{\text{'}}& \mathbf{A}\mathbf{\text{'}}\mathbf{A}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\mathbf{B}& \mathbf{A}\mathbf{\text{'}}\mathbf{A}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}\end{array}$

Finding the probability that you get two candy bars (and no money back); that you get no candy and lose your money both times; that you just get your money back both times

When $P\left(AB\right)$is subtracted from $P\left(A\right)$, the probability of getting only the bar and no money is obtained.

Find the probability of only getting the bar in both tries, that is role="math" localid="1654779029963" $\mathit{P}\left(AAB\text{'}B\text{'}\right)$.

$$\begin{gathered} \mathit{P}\left(AAB\text{'}B\text{'}\right)\mathbf{=}\left(\frac{1}{2}-\frac{1}{12}\right)\mathbf{\times }\left(\frac{1}{2}-\frac{1}{12}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{5}}{\mathbf{12}}\mathbf{\times }\frac{\mathbf{5}}{\mathbf{12}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{25}}{\mathbf{144}} \end{gathered}$$

Find the probability of not getting anything in both tries, that is role="math" localid="1654779182049" $\mathit{P}\left(A\text{'}A\text{'}B\text{'}B\text{'}\right)$.

$$\begin{gathered} \mathit{P}\left(AAB\text{'}B\text{'}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{4}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{16}} \end{gathered}$$

When $P\left(AB\right)$is subtracted from $P\left(B\right)$, the probability of getting only the money and no bar is obtained.

Find the probability of only getting the bar in both tries, that is $\mathit{P}\left(A\text{'}A\text{'}BB\right)$.

$$\begin{gathered} \mathit{P}\mathbf{(}\mathbf{A}\mathbf{\text{'}}\mathbf{A}\mathbf{\text{'}}\mathbf{BB}\mathbf{)}\mathbf{=}\left(\frac{1}{3}-\frac{1}{12}\right)\mathbf{\times }\left(\frac{1}{3}-\frac{1}{12}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}}{\mathbf{12}}\mathbf{\times }\frac{\mathbf{3}}{\mathbf{12}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{16}} \end{gathered}$$