Calculating Expected Value of Random Variables - Statistics and Probability

A spinner game defines a random variable $X$ as the number of points you score in one spin. The distribution is: - $P(X=1)=\tfrac{1}{3}$ - $P(X=4)=\tfrac{1}{3}$ - $P(X=7)=\tfrac{1}{3}$ What is the expected value of $X$ (in points)? (Note: the expected value may not be a possible outcome.)
Answer: 4 Expected value is the key concept, giving the average points you'd expect from many spins. Compute it by multiplying each X value by its probability and adding the results. Here, 1 times 1/3 is about 0.33, 4 times 1/3 is about 1.33, and 7 times 1/3 is about 2.33. Adding them yields 4 points. Over many spins, you'd average 4 points each time. Don't confuse this with the simple average of outcomes; it's weighted equally here since probabilities are equal, but generally it's a weighted average. Use the 'value times chance' approach for similar problems.
A single fair six-sided die is rolled once. Let $X$ be the number shown. The distribution is $P(X=k)=\tfrac{1}{6}$ for $k=1,2,3,4,5,6$ . What is the expected value of $X$ ? (Compute $\sum x\,P(x)$ ; the expected value may not be a possible outcome.)
Answer: 3.5 The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. For this die roll, apply it by calculating 1 × (1/6), 2 × (1/6), 3 × (1/6), 4 × (1/6), 5 × (1/6), and 6 × (1/6). Then sum these products to find the expected number shown. This value represents the long-run average result per roll. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.
A survey records the number of text messages a randomly selected person sends in an hour. Let $X$ be the number of messages. The distribution is: $P(X=0)=0.5$ , $P(X=2)=0.3$ , $P(X=4)=0.2$ . What is the expected value of $X$ (in messages)? (Compute $\sum x\,P(x)$ ; the expected value may not be a possible outcome.)
Answer: 1.4 The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. In this survey, apply it by calculating 0 × 0.5, 2 × 0.3, and 4 × 0.2. Then sum these products to find the expected number of messages. This value represents the long-run average messages per hour. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.
A raffle ticket yields winnings $X$ (in dollars) with probabilities: $P(X=-1)=\tfrac{3}{4}$ and $P(X=5)=\tfrac{1}{4}$ . (A negative value means you lose money overall.) What is the expected value of $X$ ? Give your answer in dollars.
Answer: 1/2 Expected value shows the average outcome including both wins and losses. Here, losing $1 happens with probability 3/4, while winning $5 happens with probability 1/4. We calculate: (-1)×(3/4) = -3/4 for the loss, and 5×(1/4) = 5/4 for the win. Adding these: -3/4 + 5/4 = 2/4 = 1/2. This means on average, you'd gain $0.50 per raffle ticket. Many people mistakenly think negative outcomes should be ignored, but expected value includes all possibilities. The positive result shows this raffle is actually favorable despite losing most of the time—the occasional big win outweighs frequent small losses.
A survey records how many days per week a student exercises. Let $X$ be the number of exercise days per week. The distribution is: $P(X=0)=\tfrac{1}{4}$ , $P(X=2)=\tfrac{1}{2}$ , $P(X=4)=\tfrac{1}{4}$ . What is the expected value of $X$ (in days)? (Note: the expected value may not be a possible outcome.)
Answer: 2 Expected value tells us the average number of exercise days per week across many students. We calculate it by multiplying each outcome by its probability, then summing. For X=0 days with probability 1/4, we get 0×(1/4)=0. For X=2 days with probability 1/2, we get 2×(1/2)=1. For X=4 days with probability 1/4, we get 4×(1/4)=1. Adding these: 0 + 1 + 1 = 2. This average of 2 days per week reflects that the middle value (2 days) is most likely. Don't confuse this with a simple average of (0+2+4)/3=2—here we get the same result, but only because the distribution is symmetric.
A game awards tokens. Let $X$ be the number of tokens won in one round. The distribution is: $P(X=0)=\tfrac{1}{6}$ , $P(X=3)=\tfrac{1}{2}$ , $P(X=6)=\tfrac{1}{3}$ . What is the expected value of $X$ (in tokens)? (Note: the expected value may not be a possible outcome.)
Answer: 7/2 Expected value represents the average tokens won per round over many plays. To find it, multiply each outcome by its probability and sum the results. For X=0 with probability 1/6, we get 0×(1/6)=0. For X=3 with probability 1/2, we get 3×(1/2)=3/2. For X=6 with probability 1/3, we get 6×(1/3)=2. Summing: 0 + 3/2 + 2 = 3/2 + 4/2 = 7/2. The expected value of 7/2 tokens reflects that winning 3 tokens is most likely, while the extremes are less common. This weighted average approach gives us the true long-run average, unlike simply averaging 0, 3, and 6.
A game costs $3 to play. It pays $8 with probability 0.4, otherwise $0. What is the expected net gain?
Answer: $0.20 Multiply the $8 payout by its probability 0.4 to get $3.20 expected payout. Then subtract the $3 cost to play, so the expected net gain is $0.20. The tempting $3.20 is the expected payout before subtracting the cost, not the net gain.
A spinner game defines a random variable $X$ as the prize amount (in dollars) from one spin. The distribution is: $P(X=0)=\tfrac{1}{2}$ , $P(X=4)=\tfrac{1}{4}$ , and $P(X=10)=\tfrac{1}{4}$ . What is the expected value of $X$ (in dollars)? (Compute $\sum x\,P(x)$ ; the expected value may not be a possible outcome.)
Answer: 3.5 The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. In this spinner game, apply it by calculating 0 × (1/2), 4 × (1/4), and 10 × (1/4). Then sum these products to find the expected prize amount. This value represents the long-run average prize per spin. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.
A random variable X has mean 7. What is the expected total over 100 independent observations of X?
Answer: Total 700 For each observation, the expected value is 7. Over 100 independent observations, the expected total is 100 times 7, so 700. The tempting wrong answer is "Cannot tell" because individual values vary, but the mean gives you the expected total directly.
A prize pays $10 with probability 0.25, otherwise $0. What ticket price makes expected net gain zero?
Answer: $2.50 The expected payout is 10 times 0.25 = $2.50, so a fair ticket price equals that expected value. The $10 prize amount is tempting but ignores the 75% chance you win nothing.
A fair coin is flipped twice. Define the random variable $X$ as the number of heads obtained. The distribution is: $P(X=0)=\tfrac{1}{4}$ , $P(X=1)=\tfrac{1}{2}$ , $P(X=2)=\tfrac{1}{4}$ . What is the expected value of $X$ ? (Compute $\sum x\,P(x)$ ; the expected value may not be a possible outcome.)
Answer: 1 The concept here is the expected value, which is the predicted average outcome for the random variable X over many trials. Conceptually, the formula is to multiply each possible value by its probability and then add those up. For these coin flips, apply it by calculating 0 × (1/4), 1 × (1/2), and 2 × (1/4). Then sum these products to find the expected number of heads. This value represents the long-run average number of heads per two flips. A common misconception is to take the simple average of the possible values without considering their probabilities, but we must use the weighted average based on how likely each is. For future problems, remember the strategy: think 'value × chance' for each outcome and sum them.
A bag contains cards labeled 1, 2, and 5. One card is drawn at random. Let $X$ be the number on the card. The probabilities are $P(X=1)=\tfrac{1}{2}$ , $P(X=2)=\tfrac{1}{3}$ , and $P(X=5)=\tfrac{1}{6}$ (they sum to 1). What is the expected value of $X$ ?
Answer: 2 Expected value is the key concept, serving as the predicted average value of a random variable like the number X on a drawn card. Compute it by multiplying each X value by its probability and adding the results. Here, for X=1 with probability 1/2, X=2 with 1/3, and X=5 with 1/6, the products are 0.5, about 0.67, and about 0.83. Adding them gives exactly 2. This interprets as the long-run average number if drawing many times. Don't confuse it with a simple average of the values; it's weighted by their chances. Use the 'value × chance' mindset for similar problems.
A survey records the number of minutes a student spends reading on a given day. Let $X$ be the number of minutes. The distribution is: $P(X=10)=\tfrac{1}{4}$ , $P(X=20)=\tfrac{1}{2}$ , $P(X=30)=\tfrac{1}{4}$ . What is the expected value of $X$ ? Give your answer in minutes.
Answer: 20 The concept is expected value, the average minutes spent reading per day across many surveyed students. Compute by multiplying each X by its probability and summing. For this, 10 times 1/4, 20 times 1/2, and 30 times 1/4. Adding gives 2.5 + 10 + 7.5 = 20 minutes. This interprets as an expected 20 minutes of reading on average per day. A misconception is simply averaging 10, 20, 30 to get 20 (which matches here due to symmetry), but generally, it's weighted, and asymmetry would differ. Strategy: always multiply 'value times chance' and add up.
A die-based game works like this: roll a fair six-sided die once. Let $X$ be the payout in dollars, where $X=0$ if the roll is 1–3, $X=3$ if the roll is 4–5, and $X=9$ if the roll is 6. What is the expected value of $X$ (in dollars)? (Note: the expected value may not be a possible payout.)
Answer: 2.5 Expected value represents the average payout per roll over many games. To find it, multiply each payout by its probability, then sum. For X=0 (rolls 1-3), the probability is 3/6=1/2, giving 0×(1/2)=0. For X=3 (rolls 4-5), the probability is 2/6=1/3, giving 3×(1/3)=1. For X=9 (roll 6), the probability is 1/6, giving 9×(1/6)=3/2. Summing: 0 + 1 + 3/2 = 5/2 = 2.5. This $2.50 average reflects that high payouts are rare while zero payouts are common. Remember, expected value weights outcomes by their likelihood, not just averaging the possible values.
A fair coin is flipped 4 times. For each head you win $3; for each tail you lose $2. What is the expected net gain?
Answer: $2.00 Each flip's expected gain is 3 times 1/2 plus -2 times 1/2, which equals $0.50. With 4 flips, multiply by 4 to get $2.00. The tempting mistake is $0.50, which is the expected gain for one flip only, not the full 4-flip game.
A game payout $X$ (in dollars) has distribution: $P(X=2)=\tfrac{1}{2}$ , $P(X=4)=\tfrac{1}{4}$ , $P(X=10)=\tfrac{1}{4}$ . What is the expected value of $X$ ? Give your answer in dollars.
Answer: 9/2 Expected value represents the average payout over many games. We calculate by multiplying each payout by its probability: $2 with probability 1/2 gives 2×(1/2)=1; $4 with probability 1/4 gives 4×(1/4)=1; $10 with probability 1/4 gives 10×(1/4)=5/2. Adding these: 1 + 1 + 5/2 = 2 + 5/2 = 4/2 + 5/2 = 9/2 dollars. This means you'd average $4.50 per game in the long run. Notice how the high $10 payout, despite happening only 1/4 of the time, significantly raises the average above what you might guess. Remember: multiply value by chance, then sum.
A simple game pays out dollars based on a card draw. Let $X$ be the payout from one draw. The distribution is: $P(X=0)=\tfrac{2}{5}$ , $P(X=5)=\tfrac{2}{5}$ , $P(X=10)=\tfrac{1}{5}$ . What is the expected value of $X$ (in dollars)? (Note: the expected value may not be a possible payout.)
Answer: 4 Expected value represents the average payout over many card draws. To find it, we multiply each payout by its probability and add the results. For X=0 with probability $\tfrac{2}{5}$ , we get $0 \times (\tfrac{2}{5}) = 0$ . For X=5 with probability $\tfrac{2}{5}$ , we get $5 \times (\tfrac{2}{5}) = 2$ . For X=10 with probability $\tfrac{1}{5}$ , we get $10 \times (\tfrac{1}{5}) = 2$ . Summing these products: $0 + 2 + 2 = 4$ . This means over many draws, you'd average $4 per draw. Notice how the expected value of $4$ isn't one of the possible payouts—it's the weighted average that accounts for how often each outcome occurs.
Let X be a profit in dollars taking values -$2, $1, and $5 with probabilities 0.4, 0.3, and 0.3. Find E(X).
Answer: $1.00 Multiply each profit by its probability and add: (-2)(0.4) = -0.8, (1)(0.3) = 0.3, and (5)(0.3) = 1.5. Summing gives -0.8 + 0.3 + 1.5 = 1.0, so E(X) is $1.00. The tempting $1.33 comes from simply averaging -2, 1, and 5 without using the probabilities, which is not the expected value.
A cafeteria survey defines $X$ as the number of days per week a student buys lunch at school. The distribution is: - $P(X=0)=\tfrac{1}{5}$ - $P(X=2)=\tfrac{2}{5}$ - $P(X=5)=\tfrac{2}{5}$ What is the expected value of $X$ (in days per week)? (Note: the expected value may not be a possible outcome.)
Answer: 2.8 Expected value represents average lunch-buying days per week over many students. Multiply: 0 times 1/5 is 0, 2 times 2/5 is 0.8, 5 times 2/5 is 2. Sum to 2.8 days. Long-run average is 2.8 days. Not simple average of 0,2,5 (2.33), but weighted. Strategy: value times chance, add up.
A coupon game gives a discount in dollars. Let $X$ be the discount from one play. The distribution is: $P(X=1)=\tfrac{1}{4}$ , $P(X=3)=\tfrac{1}{2}$ , $P(X=5)=\tfrac{1}{4}$ . What is the expected value of $X$ (in dollars)? (Note: the expected value may not be a possible outcome.)
Answer: 3 Expected value represents the average discount you'd receive over many plays. To find it, multiply each discount amount by its probability and sum the results. For X=1 with probability 1/4, we get 1×(1/4)=1/4. For X=3 with probability 1/2, we get 3×(1/2)=3/2. For X=5 with probability 1/4, we get 5×(1/4)=5/4. Adding these products: 1/4 + 3/2 + 5/4 = 1/4 + 6/4 + 5/4 = 12/4 = 3. The expected discount of $3 makes sense—it's exactly the middle value, which happens because the probabilities are symmetric. Remember, expected value weights outcomes by likelihood, creating a balance point for the distribution.

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