Aimathic – Expected value – Free math Worksheets

Expected value

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55143912

A game has an expected net gain of $\$0$ for the player. Does this mean the player will break even on every play? Answer yes or no and state what the expected value means.

Hints

- Distinguish one outcome from an average over many repetitions. - Think about what the word “expected” means in a probability distribution.

Solution

1. No. Expected value describes a long-run average, not the outcome of each individual play. 2. Over many plays, the average net gain per play is expected to be close to $\$0$.

Answer

No. It means the long-run average net gain per play is expected to be close to $\$0$, not that every play breaks even.

55144012

Use the equal-sector spinner shown. Find the expected prize from one spin.

Hints

- Read the possible prize values from the spinner. - Equal sectors have equal probabilities, so weight each prize by the same probability.

Solution

1. The two prizes are equally likely, so each has probability $\frac{1}{2}$. 2. The expected prize is $2\left(\frac{1}{2}\right)+6\left(\frac{1}{2}\right)=4$.

Answer

$\$4$

55142912

The table shows the net gain $X$, in dollars, from one play of a game. <table> <tr><td>$x$</td><td>$-4$</td><td>$0$</td><td>$5$</td><td>$12$</td></tr> <tr><td>$P(X=x)$</td><td>$0.20$</td><td>$0.35$</td><td>$0.30$</td><td>$0.15$</td></tr> </table> Find the expected net gain per play.

Hints

- Use every value of $X$ shown in the table. - Weight each possible gain or loss by its probability. - Check that the probabilities sum to $1$ before calculating the mean.

Solution

1. Multiply each possible gain by its probability and add the products. 2. $E(X)=(-4)(0.20)+(0)(0.35)+(5)(0.30)+(12)(0.15)=2.50$.

Answer

The expected net gain is $\$2.50$ per play.

53093112

Assume that birth months are independent and that each of the $12$ months is equally likely. a) Find the probability that at least two people in a randomly formed group of $4$ share a birth month. b) Find the smallest group size $n$ for which the probability of at least one shared birth month is greater than $50\%$. c) In a group of $5$, an opponent wins $\$5$ from you if at least two people share a birth month. You win $\$5$ if all birth months are different. Determine whether the game is fair.

Hints

- Use the complement that all birth months are different. - Check group sizes in increasing order. - A fair game has expected net gain $0$.

Solution

a) $P(\text{all different})=\frac{12\cdot11\cdot10\cdot9}{12^4}=\frac{55}{96}\approx 0.57292$. Thus $P(\text{at least one match})=1-\frac{55}{96}=\frac{41}{96}\approx 0.42708$. b) For $5$ people, $P(\text{all different})=\frac{12\cdot11\cdot10\cdot9\cdot8}{12^5}=\frac{55}{144}\approx 0.38194$, so the match probability is $\frac{89}{144}\approx 0.61806$. Since the probability for $4$ people is below $0.5$, the smallest value is $n=5$. c) Your expected net gain is $5\left(\frac{55}{144}\right)-5\left(\frac{89}{144}\right)=-\frac{85}{72}\approx -\$1.18$. The game is not fair and favors the opponent.

Answer

a) $\frac{41}{96}\approx 0.42708$ b) $n=5$ c) The game is not fair. Your expected net gain is approximately $-\$1.18$ per play.

53098712

Two experiments investigate the sum $S$ of two dice. Experiment 1: Roll two fair six-sided dice. Experiment 2: Roll one fair three-sided die labeled $1$ through $3$ and one fair nine-sided die labeled $1$ through $9$. In both experiments, the possible sums range from $2$ through $12$. a) Show that the expected sum is the same in both experiments. b) Compare the probabilities of getting a sum of exactly $7$. In which experiment is it more likely? c) In which experiment is a sum of at most $4$ more likely? Justify your answer mathematically.

Hints

- Use linearity of expected value and the mean of the labels on each fair die. - List the ordered outcomes that produce each requested sum. - Remember that the two experiments have different numbers of equally likely outcomes.

Solution

a) The expected value of a fair die labeled $1$ through $n$ is $\frac{n+1}{2}$. Thus $E(S_1)=3.5+3.5=7$, while $E(S_2)=2+5=7$. b) Experiment 1 has $36$ equally likely ordered outcomes, and $6$ produce a sum of $7$, so $P(S_1=7)=\frac{1}{6}$. Experiment 2 has $27$ outcomes and $3$ favorable ordered pairs, so $P(S_2=7)=\frac{1}{9}$. A sum of $7$ is more likely in Experiment 1. c) In Experiment 1, sums $2$, $3$, and $4$ occur in $1+2+3=6$ outcomes, so $P(S_1\le4)=\frac{1}{6}$. Experiment 2 also has $6$ favorable outcomes but only $27$ total, so $P(S_2\le4)=\frac{2}{9}$. A sum of at most $4$ is more likely in Experiment 2.

Answer

a) Both expected sums are $7$. b) Experiment 1: $\frac{1}{6}$; Experiment 2: $\frac{1}{9}$. A sum of $7$ is more likely in Experiment 1. c) Experiment 1: $\frac{1}{6}$; Experiment 2: $\frac{2}{9}$. A sum of at most $4$ is more likely in Experiment 2.

53128912

A high school class has $30$ students: $18$ girls and $12$ boys. Four students are selected at random for an anonymous survey. a) Find the probability that all four selected students are boys. b) The same selection process is carried out independently in $500$ classes with exactly the same composition. Predict how many of the selected groups will consist only of boys.

Hints

- Use combinations to count all possible four-student groups. - Count the groups formed only from the $12$ boys. - Multiply the probability for one class by $500$ to find the expected count.

Solution

a) There are $\binom{30}{4}=27{,}405$ possible groups and $\binom{12}{4}=495$ all-boy groups. Thus $P=\frac{495}{27405}=\frac{11}{609}\approx0.01806$. b) The expected number of all-boy groups in $500$ independent selections is $500\cdot\frac{11}{609}\approx9.03$, so a reasonable prediction is about $9$ classes.

Answer

a) $\frac{11}{609}\approx0.01806$, or about $1.81\%$ b) About $9$ classes

53129012

In a lottery game, $6$ distinct numbers are drawn from the integers $1$ through $45$, and order does not matter. a) Find the probability that all $6$ winning numbers are less than or equal to $15$. b) Suppose $5000$ drawings are held over many years. How many times is the event in part a) expected to occur?

Hints

- Count all unordered selections of $6$ numbers from $45$. - Count the selections made entirely from the first $15$ numbers. - Multiply the probability by $5000$ to find the expected number of occurrences.

Solution

a) There are $\binom{45}{6}=8{,}145{,}060$ possible drawings and $\binom{15}{6}=5005$ favorable drawings. Thus $P=\frac{5005}{8145060}=\frac{13}{21156}\approx0.0006145$. b) The expected number in $5000$ drawings is $5000\cdot\frac{13}{21156}\approx3.07$, so the event is expected to occur about $3$ times.

Answer

a) $\frac{13}{21156}\approx0.0006145$, or about $0.06145\%$ b) About $3$ times

55143112

A school fundraiser is comparing two prize plans for a large number of tickets. Plan A charges $\$4.00$ per ticket. A ticket wins $\$20$ with probability $0.10$, wins $\$5$ with probability $0.20$, and wins nothing otherwise. Plan B charges $\$3.50$ per ticket. A ticket wins $\$10$ with probability $0.20$, wins $\$2$ with probability $0.30$, and wins nothing otherwise. Which plan gives the fundraiser the greater expected net revenue per ticket? By how much would the expected revenues differ over $1000$ ticket sales?

Hints

- First find the expected prize payout for each plan. - Net revenue to the fundraiser is ticket price minus expected payout. - Compare the two per-ticket expected revenues before scaling to $1000$ tickets.

Solution

1. Plan A has expected payout $20(0.10)+5(0.20)=3.00$, so its expected net revenue is $4.00-3.00=\$1.00$ per ticket. 2. Plan B has expected payout $10(0.20)+2(0.30)=2.60$, so its expected net revenue is $3.50-2.60=\$0.90$ per ticket. 3. Plan A gives $\$0.10$ more expected revenue per ticket. Over $1000$ tickets, the expected difference is $1000(0.10)=\$100$.

Answer

Plan A. Its expected net revenue is $\$1.00$ per ticket versus $\$0.90$ for Plan B, an expected difference of $\$100$ over $1000$ tickets.

53098812

Two sets of dice are compared. Set A contains two fair six-sided dice. Set B contains one fair six-sided die and one weighted six-sided die. On the weighted die, each number from $1$ through $5$ has probability $0.1$, while $6$ has probability $0.5$. a) Find the probability of rolling a sum of $12$ with each set. b) Show that the probability of rolling a sum of $7$ is exactly the same for both sets. c) Find the expected sum for Set B. How much does it differ from the expected sum for Set A?

Hints

- A sum of $12$ has only one possible ordered outcome. - For a sum of $7$, factor out the constant probability from the fair die. - Use linearity of expected value.

Solution

a) For Set A, $P(S=12)=\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$. For Set B, $P(S=12)=\frac{1}{6}\cdot0.5=\frac{1}{12}$. b) Let $X$ be the fair-die result and $Y$ the weighted-die result. Then $P(X+Y=7)=\sum_{i=1}^{6}P(X=i)P(Y=7-i)=\frac{1}{6}\sum_{j=1}^{6}P(Y=j)=\frac{1}{6}$, matching Set A. c) The fair die has expected value $3.5$. The weighted die has expected value $0.1(1+2+3+4+5)+0.5(6)=4.5$. Therefore, $E(S_B)=3.5+4.5=8$, which is $1$ greater than $E(S_A)=7$.

Answer

a) Set A: $\frac{1}{36}$; Set B: $\frac{1}{12}$ b) Both probabilities equal $\frac{1}{6}$. c) $E(S_B)=8$, which is $1$ greater than $E(S_A)=7$.

55143012

A game uses a spinner with $10$ equal sections. One section pays a jackpot of $J$ dollars, two sections pay $\$8$, three sections pay $\$3$, and four sections pay $\$0$. It costs $\$5$ to play. a) Find the jackpot $J$ that makes the game fair, meaning the expected net gain to the player is $\$0$. b) A student says, “There are four listed prize amounts, so I should average $J$, $8$, $3$, and $0$ equally.” Explain why that reasoning is incorrect.

Hints

- Distinguish the prize amount from the player's net gain after paying to play. - A fair game has expected net gain $0$. - Use the number of spinner sections to determine the probability weight for each prize. - Check whether the four listed prize amounts are equally likely.

Solution

a) For a fair game, the expected prize must equal the $\$5$ cost. The expected prize is $\frac{1}{10}J+\frac{2}{10}(8)+\frac{3}{10}(3)+\frac{4}{10}(0)$. Set this equal to $5$: $\frac{J+16+9}{10}=5$, so $J+25=50$ and $J=25$. b) The four prize amounts are not equally likely. Their probabilities are $\frac{1}{10}$, $\frac{2}{10}$, $\frac{3}{10}$, and $\frac{4}{10}$, so expected value must use those probability weights.

Answer

a) $J=\$25$ b) The prize amounts cannot be averaged equally because they occur on different numbers of spinner sections and therefore have different probabilities.

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