Question

Mystery box Ms. Tyson keeps a Mystery Box in her classroom. If a student meets expectations for behavior, she or he is allowed to draw a slip of paper without looking. The slips are all of equal size, are well mixed, and have the name of a prize written on them. One of the “prizes”—extra homework—isn’t very desirable! Here is the probability model for the prizes a student can win:

a. Explain why this is a valid probability model.

b. Find the probability that a student does not win extra homework.

c. What’s the probability that a student wins candy or a homework pass?

Short answer

Part(a) It is valid probability model as it follows both conditions of probability.

Part(b) $0.95$ is the probability that a student does not win extra homework.

Part(c) $0.40$ is the probability that a student wins candy or a homework pass.

Step-by-step solution

Part(a) Step 1 : Given information

We are given a table . We need to explain validity of probability model.

Prize	Pencil	Candy	Stickers	Homework pass	Extra homework
Probability	$0.40$	$0.25$	$0.15$	$0.15$	$0.05$

Part(a) Step 2 : Simplify

As we know sum of all probabilities must be $1$and probabilities should lie between $0$and $1$ .

Sum of probabilities is $0.40+0.25+0.15+0.15+0.05=1$

Now, from the table we can say it follows both above mentioned conditions.

So, We can say the probability model is valid.

Part(b) Step 1 : Given information

We are given a table. We need to find probability that a student does not win extra homework.

Prize	Pencil	Candy	Stickers	Homework pass	Extra homework
Probability	$0.40$	$0.25$	$0.15$	$0.15$	$0.05$

Part(b) Step 2 : Simplify

Using complement rule,

$\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)=\mathrm{P}\left(\overline{\mathrm{A}}\right)=1-\mathrm{P}\left(\mathrm{A}\right)$

Probability that student wins extra homework $\mathrm{P}\left(\mathrm{E}\right)=0.05$

Now,

Probability that a student does not win extra homework

$P\left(E^c\right)=1-P\left(E\right)=1-0.05=0.95$

Hence, $0.95$ is the probability that a student does not win extra homework.

Part(c) Step 1 : Given information

We are given a table. We need to find probability that a student wins candy or a homework pass.

Prize	Pencil	Candy	Stickers	Homework pass	Extra homework
Probability	$0.40$	$0.25$	$0.15$	$0.15$	$0.05$

Part(c) Step 2 : Simplify

Using addition rule of disjoint events,

$\mathrm{P}(\mathrm{A}\mathrm{U}\mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)$

Now,

Probability that a student wins candy $P\left(C\right)=0.25$

Probability that a student wins homework pass $\mathrm{P}\left(\mathrm{H}\right)=0.15$

Probability that a student wins candy or a homework pass

$$\begin{gathered} \mathrm{P}(\mathrm{C}\mathrm{U}\mathrm{H})=\mathrm{P}\left(\mathrm{C}\right)+\mathrm{P}\left(\mathrm{H}\right) \\ \mathrm{P}(\mathrm{C}\mathrm{U}\mathrm{H})=0.25+0.15 \\ \mathrm{P}(\mathrm{C}\mathrm{U}\mathrm{H})=0.40 \end{gathered}$$

Hence, $0.40$ is the probability that a student wins candy or a homework pass.