91影视

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

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.

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

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.

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

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.