Question

A theater group holds a fund-raiser. It sells 100 raffle tickets for \(5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of \)150.

a. What are you interested in here?

b. In words, define the random variable X.

c. List the values that X may take on.

d. Construct a PDF.

e. If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle?

Short answer

(a). Net profit.

(b). $X$defines the net profit earned by copping 4 raffle tickets.

(C). $X$can take only 2 values $-20$and$130.$

(d).

$X$	$P\left(X\right)$
$-20$	$P(X=-20)=\frac{4!}{0!(100-0)!}\times {\left(\frac{1}{100}\right)}^0\times {\left(\frac{99}{100}\right)}^4=0.96$
$130$	$P(X=130)=\frac{4!}{0!(100-1)!}\times {\left(\frac{1}{100}\right)}^0\times {\left(\frac{99}{100}\right)}^3=0.04$
$TOTAL$	$1$

(e). The Expected average earning is$-14.$

Step-by-step solution

Explanation of Solution

(a).

100 raffle tickets deals for$\$5$all.

You buy 4 tickets

Prize is two passes to a Broadway show worth total$\$150.$

Interpretation:

As per the questions the most important thing is net profit, if we buy 4 raffle tickets what the net profit will is should be our main concern.

:Explanation of Solution

(b).

Given:

100 truck tickets deals for$\$5$all. You buy 4 tickets

Prize is two passes to a Broadway show worth total$\$150$.

Interpretation:

In this question,$X$is a arbitrary variable which defines the net profit earned by copping the 4 truck tickets. There are 2 possibilities$-20$when person doesn't win the prize and $130$when person wins the prize.

:Explanation of Solution

(C)

Interpretation :$X$can take only 2 values$-20$and 130 because there are 2 possibilities$-20$when person does not win the prize and 130 when person wins the prize.

:Explanation of Solution 

(D).

Interpretation: $X$can take only 2 values$-20$and 130 because there are 2 possibilities $-20$when person does not win the prize and $130$when person wins the prize. To get probabilities we have used the formula $P(X=x)=\frac{n!}{x!(n-x)!}\times p^x\times (1-p{)}^{n-x}$

:Explanation of Solution 

(E).

X	P(x)	X.P(x)
$-20$	role="math" localid="1648576197570" $P(X=-20)=\frac{4!}{0!(100-0)!}\times {\left(\frac{1}{100}\right)}^0\times {\left(\frac{99}{100}\right)}^4=0.96$	$-19.2$
$130$	role="math" localid="1648576225178" $P(X=130)=\frac{4!}{0!(100-1)!}\times {\left(\frac{1}{100}\right)}^0\times {\left(\frac{99}{100}\right)}^3=0.04$	$5.2$
TOTAL	1	$-14$

Interpretation:

The expected average earning is $-14$which we have got by using the formula.$E\left(x\right)=\sum _{}^{}[x.P(x\left)\right]$.