Question

(a) what is the expected value if you buy one ticket? fill in the table to help organize the needed information, then answer the question. round the to the nearest cent, if necessary.

Explanation:

Step1: Recall expected - value formula

The expected - value formula is $E(X)=\sum_{i}x_ip_i$, where $x_i$ is the gain and $p_i$ is the probability of that gain. Let's assume the cost of the ticket is $C$ dollars. When you lose, your gain is $-C$ (a negative gain means a loss). Let's assume the probabilities of losing, winning $\$10$, winning $\$25$, winning $\$50$, and winning $\$100$ are $p_1,p_2,p_3,p_4,p_5$ respectively, and $\sum_{i = 1}^{5}p_i=1$.

Step2: Calculate the expected - value terms

If we assume the cost of the ticket is $\$1$ (since it's not given in the problem, we need to assume a cost for the sake of calculation. Let $C = 1$). When you lose, $x_1=- 1$; when you win $\$10$, $x_2 = 10 - 1=9$; when you win $\$25$, $x_3=25 - 1 = 24$; when you win $\$50$, $x_4=50 - 1=49$; when you win $\$100$, $x_5=100 - 1 = 99$.
Let's assume the probabilities are $p_1 = 0.9$, $p_2=0.08$, $p_3=0.015$, $p_4=0.004$, $p_5=0.001$ (these are just example probabilities for illustration purposes).
$E(X)=(-1)\times0.9+9\times0.08 + 24\times0.015+49\times0.004+99\times0.001$

Step3: Perform the calculations

$E(X)=-0.9 + 0.72+0.36 + 0.196+0.099$
$E(X)=-0.9+(0.72 + 0.36+0.196 + 0.099)$
$E(X)=-0.9 + 1.375$
$E(X)=0.475\approx0.48$

Answer:

If we assume the cost of the ticket is $\$1$ and the probabilities as above, the expected value is approximately $\$0.48$. However, without the actual probabilities and the cost of the ticket given in the problem, we can't give a definite answer. If we had the full information, we would substitute the correct values into the formula $E(X)=\sum_{i}x_ip_i$ to get the exact expected value.