Sign In
Make a bar graph. Each bar should represent a prize, and the bar's height should be proportional to the probability of winning the prize.
Multiply the prizes by their corresponding probabilities. Then, sum all obtained products together.
Compare the ticket cost and the expected value.
$1.20
See solution.
Let's look at the given table.
| Distribution of Prices | |||||||||
|---|---|---|---|---|---|---|---|---|---|
| Prize | no prize | $ 20 | $ 10 | $ 5 | |||||
| Probability | 0.84 | 0.01 | 0.05 | 0.10 | |||||
We can make a bar graph of the probability distribution using the information from the table. Each bar will represent a prize, and the bar's height will be proportional to the probability of getting the prize. Here we go!
The expected value of a discrete random variable X is the weighted sum of the values of the variable. In our case, the values of the random variables are the prizes of the raffle.
E(X)= ∑ [X* P(X)]
| Distribution of Prices | |||||||||
|---|---|---|---|---|---|---|---|---|---|
| Prize, X | No prize =$0 | $ 20 | $ 10 | $ 5 | |||||
| Probability, P(X) | 0.84 | 0.01 | 0.05 | 0.10 | |||||
| X* P(X) | $0*0.84=$0 | $20* 0.01=$0.20 | $10*0.05=$ 0.50 | $5* 0.10=$0.50 | |||||
Finally, we can calculate the expected value by summing all values from the the third row. Let's do it! E(X)= $0+$0.20+$0.50+ $0.50+$0.50 =$1.20 The expected value, which is what you expect to win on average, is $1.20.
In Part B we found that the expected value — which is what a participant of the raffle can expect to gain — is $1.20. Since the price ticket is $1, which is lower than the expected value, it is worth it to participate in the raffle. However, it is not worth it to host the raffle.