Free math worksheets

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

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 \(\$0\).

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

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

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

\(\$4\)

- Compute the probability-weighted average of the two possible prizes. - Ask whether an average must always equal one of the values being averaged. - Interpret expected value over many repeated plays rather than as a guaranteed single result.

1. The expected prize is \(0\cdot\frac{1}{2}+3\cdot\frac{1}{2}=1.50\). 2. Expected value is a probability-weighted long-run average, so it does not have to be one of the possible outcomes of a single play. 3. Therefore, Maya's calculation is correct but her conclusion is not.

No. The expected prize is \(\$1.50\), and an expected value does not have to be a possible single-play outcome.

- 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.

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

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

- Think about the expected contribution to the count from one shot. - The total count is built from \(20\) identical success-or-failure contributions. - Interpret the result as a long-run average across many sets of \(20\) shots.

1. Each shot contributes either \(1\) made shot with probability \(0.35\) or \(0\) made shots with probability \(0.65\), so the expected contribution of one shot is \(0.35\). 2. Across \(20\) independent trials, \(E(X)=20\cdot0.35=7\). This is the binomial mean rule \(E(X)=np\). 3. Over many sets of \(20\) shots under these conditions, the average number made per set would approach \(7\).

\(E(X)=7\). Over many sets of \(20\) shots, the average number of made shots per set would approach \(7\).

- Find the probability that all five birth months are different before considering the game payouts. - The two game outcomes are complementary, so their probabilities must add to \(1\). - Weight each net payoff by its probability and add the results.

1. The probability that all five birth months are different is \(\frac{12\cdot11\cdot10\cdot9\cdot8}{12^5}=\frac{55}{144}\). 2. Therefore, the probability that at least two people share a birth month is \(1-\frac{55}{144}=\frac{89}{144}\). 3. The expected net gain is \(8\left(\frac{55}{144}\right)-5\left(\frac{89}{144}\right)=\frac{440-445}{144}=-\frac{5}{144}\) dollars. 4. Since the expected net gain is not \(0\), the game is not fair. It favors the opponent slightly.

The expected net gain is \(-\frac{5}{144}\) dollars, approximately \(-\$0.03\) per play. The game is not fair.

- 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.

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.

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.

- 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.

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.

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

- 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.

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.

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

- Distinguish the prize amount from the player’s net gain after paying to play. - A fair game has expected net gain \(0\). - Use the spinner to determine how likely each prize amount is. - Check whether the four listed prize amounts are equally likely.

1. For a fair game, the expected prize must equal the \(\$5\) cost. The expected prize is \(\frac{1}{10}J+\frac{2}{10}(8)+\frac{3}{10}(3)+\frac{4}{10}(0)\). 2. Set the expected prize equal to \(5\): \(\frac{J+16+9}{10}=5\), so \(J+25=50\) and \(J=25\). 3. Nikos’s reasoning is incorrect because 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.

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.

- 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.

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 \(1000(0.10)=\$100\).

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.

- 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.

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 \(E(S_A)=7\).

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 \(E(S_A)=7\).