Aimathic – Expected value – Free Math Worksheets

Expected value

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Difficulty: 1 – 5

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

1. For $4$ people, $P(\text{all different})=\frac{12\cdot11\cdot10\cdot9}{12^4}=\frac{55}{96}\approx 0.57292$. Thus, $P(\text{at least one match})=1-\frac{55}{96}=\frac{41}{96}\approx 0.42708$. 2. 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 $n=5$. 3. 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

1. 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 $E(S_2)=2+5=7$. 2. Experiment 1 has $36$ equally likely ordered outcomes, and $6$ produce a sum of $7$. Therefore, $P(S_1=7)=\frac{6}{36}=\frac{1}{6}$. Experiment 2 has $27$ equally likely ordered outcomes, and $(1, 6)$, $(2, 5)$, and $(3, 4)$ produce a sum of $7$. Thus, $P(S_2=7)=\frac{3}{27}=\frac{1}{9}$. A sum of $7$ is more likely in Experiment 1. 3. In Experiment 1, the sums $2$, $3$, and $4$ occur in $1+2+3=6$ outcomes, so $P(S_1\le 4)=\frac{6}{36}=\frac{1}{6}$. Experiment 2 also has $6$ favorable outcomes, but only $27$ total outcomes, so $P(S_2\le 4)=\frac{6}{27}=\frac{2}{9}$. 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

- Decide whether the selection is made with or without replacement. - 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

1. The total number of four-student groups is $\binom{30}{4}=27{,}405$. 2. The number of groups containing only boys is $\binom{12}{4}=495$. Therefore, $p=\frac{495}{27405}=\frac{11}{609}\approx 0.01806$. 3. Over $500$ independent selections, the expected number of all-boy groups is $500\cdot\frac{11}{609}\approx 9.03$. A reasonable prediction is about $9$ classes.

Answer

a) $\frac{11}{609}\approx 0.01806$, 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. - Divide the favorable count by the total count. - Multiply the probability by $5000$ to find the expected number of occurrences.

Solution

1. The total number of possible drawings is $\binom{45}{6}=8{,}145{,}060$. 2. For all winning numbers to be at most $15$, all $6$ numbers must be chosen from $1$ through $15$. There are $\binom{15}{6}=5005$ favorable drawings. 3. Therefore, $P=\frac{5005}{8145060}=\frac{13}{21156}\approx 0.0006145$. 4. The expected number in $5000$ drawings is $5000\cdot\frac{13}{21156}\approx 3.07$. Thus, the event is expected to occur about $3$ times.

Answer

a) $\frac{13}{21156}\approx 0.0006145$, or about $0.06145\%$ b) About $3$ times

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

1. For Set A, a sum of $12$ requires two sixes, so $P(S=12)=\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$. For Set B, $P(S=12)=\frac{1}{6}\cdot 0.5=\frac{1}{12}$. 2. Let $X$ be the fair-die result and $Y$ the weighted-die result. Then $P(X+Y=7)=\sum_{i=1}^{6}P(X=i)P(Y=7-i)=\frac{1}{6}\sum_{j=1}^{6}P(Y=j)=\frac{1}{6}$. This matches Set A. 3. The fair die has expected value $3.5$. The weighted die has expected value $0.1\cdot(1+2+3+4+5)+0.5\cdot6=4.5$. 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 $E(S_A)=7$.

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