- 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.

- Would a morning with six repairs contribute differently to the average than a morning with two repairs? - How can the probabilities help you account for mornings with different workloads?

1. Each possible number of repairs contributes to the expected count according to how likely it is. 2. Add the weighted counts: \(E(X)=0\cdot\frac{1}{2}+2\cdot\frac{1}{4}+6\cdot\frac{1}{4}=2\). 3. The expected number is \(2\) flat tires fixed per morning.

\(2\) flat tires fixed per morning on average.

- Which outcome occurs on more sections? - How does the rare larger prize affect the average? - Can an average lie between the only two possible prizes?

1. The most likely prize is \(\$0\), with probability \(\frac{3}{4}\). 2. The expected prize is \(E(X)=0\cdot\frac{3}{4}+12\cdot\frac{1}{4}=3\), so \(\$3\). 3. The expected value is a long-run average, not a prize that must be possible on one spin.

Most likely prize: \(\$0\). Expected prize: \(\$3\). The expectation is an average across many plays, not the usual single result.

- The ticket pays either a prize or nothing. What would you expect to receive before paying for it? - Check whether your final amount represents money won or the change in money after buying a ticket.

1. The expected prize before the ticket cost is \(12\cdot\frac{1}{4}+0\cdot\frac{3}{4}=3\) dollars. 2. Subtract the ticket price: \(3-4=-1\) dollars. 3. The expected net gain is a loss of \(\$1\) per ticket.

\(-\$1\), an expected loss of \(\$1\) per ticket.

- 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 \(7\).

- How is an expected payout calculated from possible payouts and their chances? - Use the given average to form an equation with \(q\). - Check that the resulting probabilities add to one.

1. Write the expected value: \(0(0.5)+5q+10(0.5-q)=4\). 2. Simplify to \(5-5q=4\), so \(q=0.2\). 3. The probability of \(\$10\) is \(0.5-0.2=0.3\). Both probabilities are valid and the three sum to \(1\).

\(q=0.2\); \(P(\$10)=0.3\).

- What is the average score from one round? - How do expected scores add across several rounds? - What actual number of successful rounds gives fewer points than expected? What about more?

1. The expected score in one round is \(15\cdot0.2+0\cdot0.8=3\) points. 2. Over \(40\) rounds, the expected total is \(40\cdot3=120\) points. 3. Scoring in \(7\) rounds would give \(7\cdot15=105\) points, below the expectation. Scoring in \(9\) rounds would give \(9\cdot15=135\) points, above it. Both are possible. Expected value describes an average over many sets of rounds, not a guaranteed result.

Expected total: \(120\) points. Possible actual totals include \(105\) and \(135\) points, depending on how many rounds score points.

- 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.

- Do the three prizes have the same chance of being won? - How can you combine each prize and its probability to find a long-run average? - Keep the cost of playing separate from the prize before deciding whether the game favors the player. - Compare the possible results of one spin with the average across many spins.

1. The probabilities total \(\frac{1}{2}+\frac{3}{10}+\frac{1}{5}=1\). Weight each prize by its probability: \(E(X)=0\cdot\frac{1}{2}+5\cdot\frac{3}{10}+15\cdot\frac{1}{5}=4.50\) dollars. 2. Subtract the cost of one spin: \(\$4.50-\$5.00=-\$0.50\). The player has an expected loss of \(\$0.50\) per spin. 3. A single spin gives a net result of \(-\$5\), \(\$0\), or \(\$10\), not \(-\$0.50\). Expected value describes the average result over many spins.

a) \(\$4.50\) b) Expected loss of \(\$0.50\) per spin. c) The actual net result of one spin is \(-\$5\), \(\$0\), or \(\$10\); expected value is a long-run average, not a guaranteed outcome.

- Think about the average result on each die before looking at the sum. - Which pairs of die faces can produce a particular sum? - Are there equally many possible pairs in the two experiments?

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-token groups. - Count the groups made only from the red tokens. - How likely is each of the two possible award amounts? - Would a simple average be reasonable if one award is much less likely?

a) There are \(\binom{30}{4}=27{,}405\) possible groups and \(\binom{12}{4}=495\) all-red groups. Thus \(P=\frac{495}{27405}=\frac{11}{609}\approx0.01806\). b) The possible payouts are \(\$20\) with probability \(\frac{11}{609}\) and \(\$2\) with probability \(\frac{598}{609}\). Thus the expected payout is \(20\cdot\frac{11}{609}+2\cdot\frac{598}{609}=\frac{1416}{609}=\frac{472}{203}\approx\$2.33\). c) The payouts are not equally likely. Averaging \(\$20\) and \(\$2\) gives them equal weight and ignores their probabilities.

a) \(\frac{11}{609}\approx0.01806\), or about \(1.81\%\) b) \(\frac{472}{203}\) dollars, approximately \(\$2.33\) c) The award amounts have different probabilities, so their ordinary average is not the expected award.

- Which two disjoint events lead to a cash prize? - Count how many drawn groups meet each event, and compare with all possible groups. - In a fair game, what should the average net gain be after paying for the ticket?

1. There are \(\binom{45}{6}=8{,}145{,}060\) equally likely drawings. All six numbers are at most \(15\) in \(\binom{15}{6}=5005\) cases, so \(P(\$1000)=\frac{5005}{8145060}\). Exactly five are at most \(15\) in \(\binom{15}{5}\binom{30}{1}=90{,}090\) cases, so \(P(\$100)=\frac{90090}{8145060}\). 2. The expected prize is \(1000\cdot\frac{5005}{8145060}+100\cdot\frac{90090}{8145060}=\frac{14014000}{8145060}\approx\$1.72\). After the ticket price, expected net gain is approximately \(-\$0.28\). It is not fair to the player because the expected net gain is negative.

a) \(P(\$1000)=\frac{5005}{8145060}\); \(P(\$100)=\frac{90090}{8145060}\) b) Expected net gain approximately \(-\$0.28\) per ticket; not fair to the player.

- 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.

- Could the player get the same prize on every play even though the two boxes hold different amounts? - Could a different placement rule change the chance of receiving each amount? - Compare the expected payout under each rule with the ticket price.

1. One rule puts \(\$0\) in A on every play. The player always receives \(\$0\), so the expected net gain is \(0-5=-5\) dollars. 2. Another rule shuffles the two prizes fairly before every play, independently of the player's fixed choice. Then each prize has probability \(\frac{1}{2}\), giving expected net gain \(0\cdot\frac{1}{2}+10\cdot\frac{1}{2}-5=0\) dollars. 3. Both rules use the same prize amounts but produce different expected gains, so the fairness claim does not follow from the two amounts alone.

For example, always putting \(\$0\) in A gives expected net gain \(-\$5\). Shuffling the prizes fairly before each play gives \(\$0\). The prize amounts alone do not make the game fair.

- Which pairs of faces can produce the sums in parts a) and b)? - How do the probabilities of those pairs differ between the two sets? - Can you find the mean for each die before considering their sum?

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 \(\frac{1}{6}\). For each possible result of the weighted die, exactly one result of the fair die makes the sum \(7\). Those possibilities have total probability \(\frac{1}{6}\). c) \(E(S_B)=8\), which is \(1\) greater than \(E(S_A)=7\).

- After one card is seen, which possible prizes remain available? - For a particular first prize, which choice has a better average return after any extra cost? - Could the best decision change depending on the first card? - How likely is each possible first card before the student starts playing?

1. The original \(\$4\) fee is paid either way. After a first prize of \(\$0\), redrawing gives an expected payout of \(\frac{2+6+8}{3}=\frac{16}{3}\), with an extra \(\$1\) fee. Drawing again is better than keeping \(\$0\). 2. After \(\$2\), the expected redraw payout is \(\frac{0+6+8}{3}=\frac{14}{3}\). After the extra fee this is worth \(\frac{11}{3}\), which is greater than keeping \(\$2\). 3. After \(\$6\), redrawing gives \(\frac{0+2+8}{3}-1=\frac{7}{3}\), less than keeping \(\$6\). After \(\$8\), redrawing gives \(\frac{0+2+6}{3}-1=\frac{5}{3}\), less than keeping \(\$8\). Keep either of these prizes. 4. After the original \(\$4\) fee, the four possible first cards give conditional expected net gains of \(\frac{1}{3}\), \(-\frac{1}{3}\), \(2\), and \(4\) dollars under the better choices. 5. Each first card has probability \(\frac{1}{4}\). The expected net gain is \(\frac{1}{4}\left(\frac{1}{3}-\frac{1}{3}+2+4\right)=\frac{3}{2}\) dollars.

a) Draw again after \(\$0\) or \(\$2\); keep \(\$6\) or \(\$8\). b) Expected net gain: \(\$1.50\) per play.