5) A contractor is considering a sale that promises a profit of
3000 with a probability of 0.3. What is the expected profit? A)
23,800 C)
21,700 16) (5 pts) If X is a hypergeometric random variable, compute P(X=3) for N = 8, K = 3, and n = 5. A) 0.714 B) 0.18 C) 0.179 D) Other: List the sample space and tell whether the events are equally likely. 17) Toss a fair coin five times; record the number of heads. A) {0, 1, 2, 3, 4, 5}, equally likely B) {1, 2, 3, 4, 5}, not equally likely C) {0, 1, 2, 3, 4, 5}, not equally likely D) {1, 2, 3, 4, 5}, equally likely E) {H, T}, equally likely
The Correct Answer and Explanation is:
Question 5: Expected Profit
Problem:
A contractor is considering a sale that promises a profit of 3000 with a probability of 0.3. What is the expected profit?
Answer:
To calculate the expected profit, we use the formula for expected value:E=(Profit)×(Probability)+(0)×(1−Probability)E = (Profit) \times (Probability) + (0) \times (1 – Probability)E=(Profit)×(Probability)+(0)×(1−Probability)E=3000×0.3+0×0.7=900E = 3000 \times 0.3 + 0 \times 0.7 = 900E=3000×0.3+0×0.7=900
Correct answer: $900 (None of the answer choices listed match this. Possibly an error in options.)
Explanation:
Expected value gives the average profit the contractor can expect over many similar situations. In this case, there is only a 30 percent chance of gaining $3000 and a 70 percent chance of gaining nothing. Multiplying the profit by its probability gives an expected value of $900. This is a theoretical average; the contractor either earns $3000 or nothing in a single trial, but over many trials, the average would trend toward $900.
Question 16: Hypergeometric Probability
Given:
- N=8N = 8N=8: total items
- K=3K = 3K=3: number of successes in the population
- n=5n = 5n=5: number of draws
- Find P(X=3)P(X = 3)P(X=3)
Use the hypergeometric probability formula:P(X=k)=(Kk)(N−Kn−k)(Nn)P(X = k) = \frac{{\binom{K}{k} \binom{N – K}{n – k}}}{{\binom{N}{n}}}P(X=k)=(nN)(kK)(n−kN−K)P(X=3)=(33)(52)(85)=1×1056=1056≈0.179P(X = 3) = \frac{{\binom{3}{3} \binom{5}{2}}}{{\binom{8}{5}}} = \frac{{1 \times 10}}{{56}} = \frac{10}{56} \approx 0.179P(X=3)=(58)(33)(25)=561×10=5610≈0.179
Correct answer: C) 0.179
Explanation:
This problem models sampling without replacement, which is exactly what the hypergeometric distribution describes. We want the probability of drawing 3 “successes” (items of interest) when we randomly pick 5 items out of a population of 8 that contains 3 successes. Calculating the combinations involved allows us to find the probability accurately.
Question 17: Sample Space for Coin Toss
Situation:
Toss a fair coin five times; record the number of heads.
Sample space: {0, 1, 2, 3, 4, 5}
Correct answer: C) {0, 1, 2, 3, 4, 5}, not equally likely
Explanation:
Each individual outcome (like HHTHT) is equally likely, but the number of heads is not equally distributed. For example, there is only 1 way to get all 5 heads, but there are 10 different ways to get exactly 2 heads. Therefore, the possible numbers of heads range from 0 to 5, but each is not equally likely because some counts of heads occur in more outcome combinations than others.
Summary of Correct Answers:
- Q5: Expected profit = $900 (not listed)
- Q16: C) 0.179
- Q17: C) {0, 1, 2, 3, 4, 5}, not equally likely
