a charity is holding a raffle to raise money. there is one car worth $30000 and five $100 gift - cards being…

a charity is holding a raffle to raise money. there is one car worth $30000 and five $100 gift - cards being raffled off. each ticket costs $20, and there are a total of 5000 tickets being sold. which equation correctly depicts the calculation of the expected value for a ticket?\n$30000(\\frac{1}{5000}) + 100(\\frac{1}{1000})+(-20)(\\frac{2497}{2500}) = e(x)$\n$29980(\\frac{1}{5000})+80(\\frac{1}{1000})+(-20)(\\frac{2497}{2500}) = e(x)$\n$30000(\\frac{1}{5000}) + 100(\\frac{1}{1000}) = e(x)$\n$29980(\\frac{1}{5000})+80(\\frac{1}{1000}) = e(x)$

a charity is holding a raffle to raise money. there is one car worth $30000 and five $100 gift - cards being raffled off. each ticket costs $20, and there are a total of 5000 tickets being sold. which equation correctly depicts the calculation of the expected value for a ticket?\n$30000(\\frac{1}{5000}) + 100(\\frac{1}{1000})+(-20)(\\frac{2497}{2500}) = e(x)$\n$29980(\\frac{1}{5000})+80(\\frac{1}{1000})+(-20)(\\frac{2497}{2500}) = e(x)$\n$30000(\\frac{1}{5000}) + 100(\\frac{1}{1000}) = e(x)$\n$29980(\\frac{1}{5000})+80(\\frac{1}{1000}) = e(x)$

Answer

Explanation:

Step1: Calculate probabilities

The probability of winning the car is $\frac{1}{5000}$ since there is 1 car and 5000 tickets. The probability of winning a gift - card is $\frac{5}{5000}=\frac{1}{1000}$ as there are 5 gift - cards and 5000 tickets. The probability of winning nothing is $\frac{5000-(1 + 5)}{5000}=\frac{4994}{5000}=\frac{2497}{2500}$.

Step2: Calculate net winnings

If you win the car worth $$30000$, but you paid $$20$ for the ticket, the net winning is $30000 - 20=29980$. If you win a $$100$ gift - card and paid $$20$ for the ticket, the net winning is $100 - 20 = 80$. If you win nothing, you lose the $$20$ you paid for the ticket, so the net winning is $-20$.

Step3: Use expected - value formula

The expected - value formula is $E(X)=\sum_{i}x_ip_i$, where $x_i$ are the possible outcomes and $p_i$ are their corresponding probabilities. So $E(X)=29980\times\frac{1}{5000}+80\times\frac{1}{1000}+(-20)\times\frac{2497}{2500}$.

Answer:

$29980\left(\frac{1}{5000}\right)+80\left(\frac{1}{1000}\right)+(-20)\left(\frac{2497}{2500}\right)=E(X)$