Question

4. Suppose that you receive a movie-rental bill for the month that is much higher thanit usually is. Currently you are paying $3.99 for each movie you rent. Switching to asubscription would allow you to watch unlimited movies for only $7.99 per month. However,during a normal month you don't have much time to sit and watch movies. You do not reallywant to waste your money on a monthly subscription. You decide to check your onlinebilling statements and make a probability distribution for the number of movies you mightwatch each month.The results are in the following table:Number of Movies, X 0 1 2 3 4 5Probability, P(X) 0.10 0.15 ? 0.35 0.14 0.13a) What is the probability you will watch 2 movies next month, i.e., P(X=2)?b) What is the probability that you will watch more than 2 movies next month, i.e., P(X<4) orP(X<=3)?c) How much would you expect to spend, on a per-month basis, should you continue to pay foreach movie separately? Explain your answer

Answer 1

For the given table representing the probability distribution, the probability of watching 2 movies is unknown.

The sum of all probabilities should be equal to 1. We can use that to calculate the unknown probability:

Adding all probabilities and equating to 1:

`0.1+0.15+P(x=2)+0.35+0.14+0.13=1`

Solving for P(x=2)

`\begin{gathered} 0.87+P(x=2)=1 \\ P(x=2)=1-0.87 \\ P(x=2)=0.13 \end{gathered}`

Then: A. The probability of watching 2 movies next month is 0.13.

The complete table of probability distribution will look like this:

xP(x)

00.1

10.15

20.13

30.35

40.14

50.13

To calculate the probability of watching more than two movies we need to add the probabilities of watching 3, 4 or 5 movies. Those should be added because those are the cases where more than 2 movies are watched.

`\begin{gathered} P(x>2)=P(x=3)+P(x=4)+P(x=5) \\ P(x>2)=0.35+0.14+0.13 \\ P(x>2)=0.62 \end{gathered}`

Then, B. The probability of watching more than 2 movies is 0.62.

The probability of watching 2 movies is equivalent to P(x>2) or P(x>=3).

On the other hand, to calculate P(X<4) or P(X<=3) we need to add the probabilities of watching 3 movies or less. That is, probabilities of watching 0, 1, 2 or 3:

`\begin{gathered} P(x<4)\text{ or }P(x\le3)=P(x=0)+P(x=1)+P(x=2)+P(x=3) \\ P(x<4)\text{ or }P(x\le3)=0.1+0.15+0.13+0.35 \\ P(x<4)\text{ or }P(x\le3)=0.73 \end{gathered}`

The probability of watching 3 movies or less next month is 0.73.

To estimate how much we would expect to spend per month if we pay for each movie sepparately we need to calculate the expected value of movies per month.

We can estimate that with the probability distribution given in the table.

The expected value is the sum of the products between each event and their probabilities:

`\text{Expected Value}=\sum ^{}_{}x\cdot P(x)`

Let's call EV the expected value:

`\begin{gathered} EV=(0\cdot0.1)+(1\cdot0.15)+(2\cdot0.13)+(3\cdot0.35)+(4\cdot0.14)+(5\cdot0.13) \\ EV=2.67 \end{gathered}`

Then, we should expect to watch about 2.67 movies per month, on average.

I each individual movie costs $3.99, then, the total expenses per month will be:

`2.67\cdot3.99\approx10.65`

Then, C. According to the given probability distribution, we should expect to watch about 2.67 movies per month, on average, and spend in total $10.65 per month. We would be spending more than we would if we selected the unlimited movies plan which costs only $7.99 per month, then, it would be wise to decide to change our subscription to that plan.