Question

homework 34: problem 3 (1 point) mark draws one card from a standard deck of 52. he receives $0.75 for the king of clubs, $0.40 for any other club, and $0.65 for any other king. how much should he be willing to pay for one draw? you have attempted this problem 0 times. you have unlimited attempts remaining. preview my answers submit answers show past answers email instructor webwork © 1996 - 2025 | theme: math4 | www - version: 2.20 | pg_version: 2.7.0 page generated september 24, 2025, 2:22:50 pm cdt the webwork project

Explanation:

Step1: Calculate probabilities of each case

There is 1 king of clubs in a 52 - card deck, so $P(\text{king of clubs})=\frac{1}{52}$. There are 12 other clubs (13 clubs total minus the king of clubs), so $P(\text{other club})=\frac{12}{52}$. There are 3 other kings (4 kings total minus the king of clubs), so $P(\text{other king})=\frac{3}{52}$. The probability of getting a non - club non - king card is $P(\text{non - club non - king})=\frac{52-(1 + 12+3)}{52}=\frac{36}{52}$.

Step2: Calculate expected value

The expected value $E(X)$ is calculated by multiplying each outcome value by its probability and summing them up. $E(X)=0.75\times\frac{1}{52}+0.40\times\frac{12}{52}+0.65\times\frac{3}{52}+0\times\frac{36}{52}$. First, calculate each product: $0.75\times\frac{1}{52}=\frac{0.75}{52}$, $0.40\times\frac{12}{52}=\frac{4.8}{52}$, $0.65\times\frac{3}{52}=\frac{1.95}{52}$. Then sum them: $E(X)=\frac{0.75 + 4.8+1.95}{52}=\frac{7.5}{52}\approx0.144$.

Answer:

$0.144$