Discrete Random Variables - IB Mathematics Applications and Interpretation

1. A discrete random variable $X$ has the probability distribution $P(X=1)=p$, $P(X=2)=q$, and $P(X=3)=0.3$. Given that the expected value $E(X) = 1.9$, what is the value of $p$?
   Answer: 0.4 We have two unknowns, $p$ and $q$, so we need to set up a system of two linear equations. Equation 1: The sum of probabilities is 1. $$p + q + 0.3 = 1 \implies p + q = 0.7$$ Equation 2: The expected value is 1.9. $$E(X) = 1(p) + 2(q) + 3(0.3) = 1.9$$ $$p + 2q + 0.9 = 1.9 \implies p + 2q = 1.0$$ Now we solve the system: 1) $p + q = 0.7$ 2) $p + 2q = 1.0$ Subtracting equation (1) from equation (2): $(p + 2q) - (p + q) = 1.0 - 0.7 \implies q = 0.3$ Substitute $q=0.3$ back into equation (1): $p + 0.3 = 0.7 \implies p = 0.4$ Verification: $E(X) = 1(0.4) + 2(0.3) + 3(0.3) = 0.4 + 0.6 + 0.9 = 1.9$ ✓
2. A lottery ticket has the following probability distribution for the prize, $X$. $P(X=0)=0.6$, $P(X=10)=0.3$, and $P(X=k)=0.1$. The expected prize value of the ticket is $8. Find the value of the prize $k$.
   Answer: 50 The expected value is given by the formula $E(X) = \sum x \cdot P(X=x)$. We are given $E(X) = 8$. $$E(X) = (0)(0.6) + (10)(0.3) + (k)(0.1)$$ $$8 = 0 + 3 + 0.1k$$ $$8 = 3 + 0.1k$$ Subtract 3 from both sides: $$5 = 0.1k$$ Divide by 0.1: $$k = \frac{5}{0.1} = 50$$ Distractor C (80) results from ignoring the $10 prize ($8 = 0.1k$). Distractor D (110) results from an algebraic error ($8+3=0.1k$).
3. A random variable $X$ follows the probability distribution $P(X=x) = \frac{x}{10}$ for $x \in \{1, 2, 3, 4\}$. Let $Y = 10 - 2X$. Find $E(Y)$.
   Answer: 4.0 First, we need to calculate $E(X)$. The probabilities are $P(X=1)=0.1, P(X=2)=0.2, P(X=3)=0.3, P(X=4)=0.4$. This is a valid distribution as $0.1+0.2+0.3+0.4=1.0$. $$E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$$ Now, we find $E(Y)$ using the linearity of expectation: $E(Y) = E(10 - 2X) = 10 - 2E(X)$. $$E(Y) = 10 - 2(3.0) = 10 - 6 = 4.0$$ Distractor A (3.0) is the value of $E(X)$. Distractor D (7.0) results from calculating $10-E(X)$. Distractor C (6.0) is the value of $2E(X)$.
4. The probability distribution of a discrete random variable $X$ is given by $P(X=0)=0.1$, $P(X=1)=0.3$, $P(X=2)=0.4$, and $P(X=3)=0.2$. Find the variance of $X$.
   Answer: 0.81 First, calculate the expected value, $E(X)$. $$E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7$$ Next, calculate $E(X^2)$. $$E(X^2) = 0^2(0.1) + 1^2(0.3) + 2^2(0.4) + 3^2(0.2) = 0 + 0.3 + 1.6 + 1.8 = 3.7$$ Finally, calculate the variance, $Var(X) = E(X^2) - [E(X)]^2$. $$Var(X) = 3.7 - (1.7)^2 = 3.7 - 2.89 = 0.81$$ Distractor B (0.90) is the standard deviation ($\sqrt{0.81}$). Distractor C (1.70) is $E(X)$. Distractor D (3.70) is $E(X^2)$.
5. The number of typos, $X$, on a randomly chosen page of a book is a discrete random variable with $E(X) = 2.5$. An editor charges a fee based on the square of the number of typos, given by the formula $C = 4X^2$. Which of the following statements must be true about the expected fee, $E(C)$?
   Answer: $E(C) > 25$ When you encounter questions about the expected value of transformed random variables, the key concept is understanding how expectation behaves under non-linear transformations. This tests whether you know that $E[g(X)] \neq g(E[X])$ when $g$ is non-linear. Given that $E(X) = 2.5$ and $C = 4X^2$, we need $E(C) = E(4X^2) = 4E(X^2)$. The crucial insight is that $E(X^2) \neq [E(X)]^2$ for any random variable with non-zero variance. Since $X$ represents typos on a page (a count that naturally varies), it must have positive variance. Using the variance formula: $\text{Var}(X) = E(X^2) - [E(X)]^2$, we can rearrange to get $E(X^2) = \text{Var}(X) + [E(X)]^2$. Since $\text{Var}(X) > 0$, we have $E(X^2) > [E(X)]^2 = (2.5)^2 = 6.25$. Therefore, $E(C) = 4E(X^2) > 4(6.25) = 25$. Option A incorrectly assumes $E(C) = 4[E(X)]^2 = 4(2.5)^2 = 25$, falling into the common trap of treating expectation as if it's linear for all functions. Option B suggests we need the full distribution, but we only need to know that variance is positive. Option C incorrectly claims the expected fee is less than 25, contradicting our analysis. Remember: for any non-linear transformation of a random variable, you cannot simply plug the expected value into the function. Jensen's inequality tells us that for convex functions like $x^2$, the expectation of the transformed variable exceeds the transformation of the expectation.
6. A game involves rolling a fair six-sided die. The payout, $X$, is the square of the number rolled. What is the expected payout of this game?
   Answer: 15.17 Let $Y$ be the number rolled on a fair six-sided die. The possible outcomes for $Y$ are {1, 2, 3, 4, 5, 6}, each with probability $1/6$. The payout, $X$, is given by $X = Y^2$. The question asks for $E(X)$, which is $E(Y^2)$. We need to calculate the expected value of the squared outcomes. $$E(Y^2) = \sum y^2 P(Y=y)$$ $$E(Y^2) = 1^2(\frac{1}{6}) + 2^2(\frac{1}{6}) + 3^2(\frac{1}{6}) + 4^2(\frac{1}{6}) + 5^2(\frac{1}{6}) + 6^2(\frac{1}{6})$$ $$E(Y^2) = \frac{1}{6} (1 + 4 + 9 + 16 + 25 + 36) = \frac{1}{6} (91)$$ $$E(Y^2) = \frac{91}{6} \approx 15.1666...$$ This rounds to 15.17. Distractor A (3.5) is $E(Y)$, the expected value of the roll itself. Distractor B (12.25) is $[E(Y)]^2$, a common error. Distractor D (21) is the sum of the outcomes, not the expected value.
7. A discrete random variable $X$ has the probability distribution given by $P(X=10)=0.4$, $P(X=20)=0.3$, $P(X=30)=k$, and $P(X=40)=0.1$. What is the expected value of $X$?
   Answer: 20.0 First, find the value of $k$ using the property that the sum of probabilities for a discrete random variable is 1. $0.4 + 0.3 + k + 0.1 = 1$, which gives $0.8 + k = 1$, so $k = 0.2$. Next, calculate the expected value, $E(X)$, using the formula $E(X) = \sum x \cdot P(X=x)$. $$E(X) = (10)(0.4) + (20)(0.3) + (30)(0.2) + (40)(0.1)$$ $$E(X) = 4 + 6 + 6 + 4 = 20.0$$ Distractor A (17.0) results from incorrectly assuming $k=0.1$. Distractor C (22.5) is the median value, not the mean. Distractor D (25.0) is the simple arithmetic mean of the outcomes $(10+20+30+40)/4$, which ignores the probabilities.
8. The score, $X$, in a game has the probability distribution $P(X=1)=0.5$, $P(X=2)=0.3$, $P(X=3)=0.2$. A player's prize money, $M$, is calculated using the formula $M = 5X^2 + 10$. What is the expected prize money, $E(M)$?
   Answer: 27.50 To find the expected prize money, we use the property of expectation: $E(M) = E(5X^2 + 10) = 5E(X^2) + 10$. First, we need to calculate $E(X^2)$. $$E(X^2) = 1^2(0.5) + 2^2(0.3) + 3^2(0.2)$$ $$E(X^2) = 1(0.5) + 4(0.3) + 9(0.2) = 0.5 + 1.2 + 1.8 = 3.5$$ Now substitute this value back into the equation for $E(M)$. $$E(M) = 5(3.5) + 10 = 17.5 + 10 = 27.5$$ The expected prize money is $27.50. Distractor B (24.45) is a common error from calculating $5[E(X)]^2 + 10$ instead of $5E(X^2) + 10$. Distractor A (17.50) forgets to add the constant 10.
9. A discrete random variable $X$ has $E(X) = 5$ and $E(X^2) = 29$. A new random variable $Y$ is defined as $Y = 2X - 3$. What is the variance of $Y$?
   Answer: 16 This is a multi-step problem. First, we need to find the variance of $X$. Then, we use the properties of variance to find the variance of $Y$. Step 1: Find $Var(X)$. $Var(X) = E(X^2) - [E(X)]^2$ $Var(X) = 29 - (5)^2 = 29 - 25 = 4$ Step 2: Find $Var(Y)$ using the property $Var(aX + b) = a^2 Var(X)$. For $Y = 2X - 3$, we have $a=2$ and $b=-3$. $Var(Y) = 2^2 \cdot Var(X) = 4 \cdot 4 = 16$. Distractor A (4) is the variance of $X$. Distractor B (11) is from an incorrect calculation for Var(Y) as $2Var(X)-3$. Distractor D (55) results from calculating $E(Y^2)$ without finding variance.
10. In a game, a player's score $X$ has the following probability distribution: $P(X=-2)=0.1$, $P(X=1)=0.5$, $P(X=5)=0.4$. Find the standard deviation of $X$, correct to three significant figures.
    Answer: 2.37 First, find the expected value, $E(X)$. $$E(X) = (-2)(0.1) + (1)(0.5) + (5)(0.4) = -0.2 + 0.5 + 2.0 = 2.3$$ Next, find $E(X^2)$. $$E(X^2) = (-2)^2(0.1) + (1)^2(0.5) + (5)^2(0.4) = 4(0.1) + 1(0.5) + 25(0.4) = 0.4 + 0.5 + 10.0 = 10.9$$ Then, find the variance, $Var(X) = E(X^2) - [E(X)]^2$. $$Var(X) = 10.9 - (2.3)^2 = 10.9 - 5.29 = 5.61$$ Finally, the standard deviation is the square root of the variance: $$\sigma = \sqrt{Var(X)} = \sqrt{5.61} \approx 2.3685...$$ Rounding to three significant figures gives 2.37. Distractor D (5.61) is the variance. Distractor B (2.30) is the expected value. Distractor A (2.24) results from an error in calculating $E(X^2)$ where $(-2)^2$ is taken as -4.
11. An insurance company sells a policy for $250. The policy covers an accident that has a 2% probability of occurring. If the accident occurs, the company pays out $10,000. Let $X$ be the company's profit from one policy. Find the expected profit, $E(X)$, for the company per policy sold.
    Answer: $50 When you encounter expected value problems in business contexts, you're calculating the long-run average outcome by weighing each possible result by its probability. To find the company's expected profit per policy, you need to identify all possible outcomes and their probabilities. There are two scenarios: either the accident occurs (2% chance) or it doesn't (98% chance). If no accident occurs (98% probability): The company keeps the $250 premium and pays nothing, so profit = $250. If an accident occurs (2% probability): The company receives $250 but pays out $10,000, so profit = $250 - $10,000 = -$9,750. The expected profit is: $E(X) = 0.98 × 250 + 0.02 × (-9,750) = 245 - 195 = 50$ Therefore, the answer is D) $50. Let's examine why the other options are incorrect: A) -$200 likely comes from incorrectly focusing only on the loss scenario or making calculation errors with the probabilities. B) $250 represents the premium collected, but this ignores the potential payout entirely. This would only be correct if there were no risk of claims. C) $245 results from calculating only the "no accident" scenario (0.98 × $250) while completely ignoring the negative impact of potential payouts. Study tip: In expected value problems involving insurance or gambling, always identify every possible outcome, calculate the profit/loss for each scenario, then multiply by probabilities. The key insight is that profit equals revenue minus costs, not just revenue.
12. What is the baker's expected daily profit?
    Answer: $13.50 First, determine the profit for each outcome of $X$ (number of cakes sold), given that 3 cakes are always made. The total cost is $3 \times \$5 = \$15$. The profit is Revenue - Cost. - If $X=0$, Revenue = $0. Profit = $0 - $15 = -$15. - If $X=1$, Revenue = $15. Profit = $15 - $15 = $0. - If $X=2$, Revenue = $30. Profit = $30 - $15 = $15. - If $X=3$, Revenue = $45. Profit = $45 - $15 = $30. Now, calculate the expected profit: $$E(\text{Profit}) = (-15)(0.1) + (0)(0.2) + (15)(0.4) + (30)(0.3)$$ $$E(\text{Profit}) = -1.5 + 0 + 6.0 + 9.0 = 13.5$$ The expected daily profit is $13.50. Distractor A ($12.00) results from calculation errors. Distractor C ($15.00) assumes no losses from unsold cakes. Distractor D ($16.50) overestimates the expected sales.
13. A biased coin is tossed twice. The probability of getting a Head (H) is 0.6. Let $X$ be the number of Heads obtained. Find the variance of $X$.
    Answer: 0.48 First, establish the probability distribution for $X$, the number of heads. $P(H) = 0.6$, $P(T) = 0.4$. $P(X=0) = P(TT) = (0.4)(0.4) = 0.16$ $P(X=1) = P(HT \text{ or } TH) = (0.6)(0.4) + (0.4)(0.6) = 0.24 + 0.24 = 0.48$ $P(X=2) = P(HH) = (0.6)(0.6) = 0.36$ Next, calculate $E(X)$ and $E(X^2)$. $E(X) = 0(0.16) + 1(0.48) + 2(0.36) = 0 + 0.48 + 0.72 = 1.2$ $E(X^2) = 0^2(0.16) + 1^2(0.48) + 2^2(0.36) = 0 + 0.48 + 1.44 = 1.92$ Finally, calculate the variance: $Var(X) = E(X^2) - [E(X)]^2 = 1.92 - (1.2)^2 = 1.92 - 1.44 = 0.48$. Alternatively, using the binomial distribution formula $Var(X) = np(1-p)$ for $n=2, p=0.6$, $Var(X) = 2(0.6)(0.4) = 0.48$. Distractor C is $E(X)$, and D is $E(X^2)$. Distractor B is the variance for a fair coin.
14. A machine produces bolts. The number of defective bolts per batch, $X$, is a random variable with a variance of $Var(X) = 4$. The cost, in dollars, associated with the defects is modelled by the equation $C = 10X + 50$. What is the variance of the cost, $Var(C)$?
    Answer: 400 We use the property of variance that $Var(aX + b) = a^2 Var(X)$. Here, $a = 10$ and $b = 50$. We are given $Var(X) = 4$. Therefore, $Var(C) = Var(10X + 50) = 10^2 \cdot Var(X)$. $$Var(C) = 100 \cdot 4 = 400$$ The constant $b=50$ does not affect the variance. Distractor A (90) comes from the incorrect calculation $10 \cdot 4 + 50$. Distractor C (450) from $10^2 \cdot 4 + 50$. Distractor D (1650) from $(10\cdot4+50)^2$. A common error is to calculate $a \cdot Var(X)$ instead of $a^2 \cdot Var(X)$, which would give $10 \cdot 4 = 40$.
15. A discrete random variable $X$ can take the values 1, 2, 3, and 4. Its probability distribution is given by $P(X=x) = kx$ for some constant $k$. Find the expected value of $X$.
    Answer: 3.0 First, we must find the value of the constant $k$. The sum of all probabilities must be 1. $P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$ $k(1) + k(2) + k(3) + k(4) = 1$ $k + 2k + 3k + 4k = 1$ $10k = 1 \implies k = 0.1$ Now we have the probabilities: $P(X=1)=0.1, P(X=2)=0.2, P(X=3)=0.3, P(X=4)=0.4$. Calculate the expected value, $E(X)$: $$E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4)$$ $$E(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$$ Distractor A (2.5) is the arithmetic mean of the values, ignoring probabilities. Other distractors result from calculation errors.
16. The discrete random variable $X$ has probability distribution $P(X=0)=0.6$ and $P(X=a)=0.4$, where $a>0$. The variance of $X$ is given as $Var(X) = 3.84$. Find the value of $a$.
    Answer: 4.0 First, find expressions for $E(X)$ and $E(X^2)$ in terms of $a$. $E(X) = 0(0.6) + a(0.4) = 0.4a$ $E(X^2) = 0^2(0.6) + a^2(0.4) = 0.4a^2$ Now use the variance formula: $Var(X) = E(X^2) - [E(X)]^2$. $$3.84 = 0.4a^2 - (0.4a)^2$$ $$3.84 = 0.4a^2 - 0.16a^2$$ $$3.84 = 0.24a^2$$ $$a^2 = \frac{3.84}{0.24} = 16$$ Since $a>0$, we take the positive square root: $a = \sqrt{16} = 4$. Distractor B (3.2) results from forgetting to square the 0.4 inside the bracket: $3.84 = 0.4a^2 - 0.4a^2$ which doesn't work, or some other algebra error. Distractor D (9.6) comes from $3.84/0.4$.
17. A company models its daily sales of a specific product, $X$, as a discrete random variable. After analyzing historical data, they calculate the expected value to be $E(X) = 8.2$. Which statement is the best interpretation of this expected value?
    Answer: Over a long period, the average number of products sold per day will be close to 8.2. When you encounter expected value questions, you're dealing with the fundamental concept of what happens "on average" over many repetitions of a random process. The expected value $E(X) = 8.2$ represents the theoretical mean of the distribution - the center point around which actual outcomes will cluster over time. Option D correctly captures this interpretation. Expected value tells you the long-run average: if this company tracked daily sales over hundreds or thousands of days, the mean of all those daily totals would approach 8.2. This is the law of large numbers in action. Option A reflects a common misconception about expected value. Since $X$ is discrete (you can't sell 8.2 products), the expected value doesn't have to be a possible outcome. It's a theoretical average, not a prediction for any single day. Option B incorrectly assumes the expected value guarantees outcomes above 8 on most days. Expected value tells you nothing about the probability distribution around that mean - the company could sell 0 products on many days and 20+ on others, still averaging 8.2. Option C confuses expected value with mode (the most frequent outcome). The expected value is the mean of the distribution, not necessarily the most likely single outcome. In fact, 8.2 isn't even a possible outcome for this discrete variable. Remember: expected value always describes long-term average behavior, never single-event predictions. When you see expected value questions on the IB exam, focus on the "over many trials" interpretation rather than what happens on any individual trial.
18. A discrete random variable $X$ has the following properties: $E(X) = 10$ and standard deviation $\sigma = 3$. What is the value of $E(X^2)$?
    Answer: 109 The relationship between variance, expected value, and the expectation of the square is given by $Var(X) = E(X^2) - [E(X)]^2$. We are given the standard deviation $\sigma = 3$, so the variance is $Var(X) = \sigma^2 = 3^2 = 9$. We are also given $E(X) = 10$. Substitute these values into the variance formula: $$9 = E(X^2) - (10)^2$$ $$9 = E(X^2) - 100$$ Rearrange to solve for $E(X^2)$: $$E(X^2) = 100 + 9 = 109$$ Distractor A (91) is from incorrectly subtracting the variance: $100 - 9$. Distractor B (100) is $[E(X)]^2$. Distractor C (103) is from adding the standard deviation instead of the variance: $100+3$.
19. Let $X$ be the daily high temperature in degrees Celsius in a city, with an expected value $E(X) = 15$. The temperature in degrees Fahrenheit, $Y$, is given by the formula $Y = 1.8X + 32$. What is the expected daily high temperature in degrees Fahrenheit?
    Answer: 59.0 We use the property of expected values that $E(aX + b) = aE(X) + b$. Here, $a = 1.8$ and $b = 32$. We are given $E(X) = 15$. Therefore, $E(Y) = E(1.8X + 32) = 1.8 \cdot E(X) + 32$. $$E(Y) = 1.8(15) + 32 = 27 + 32 = 59.0$$ Distractor A (27.0) forgets to add the constant 32. Distractor B (47.0) forgets to multiply by the coefficient 1.8. Distractor D (80.6) incorrectly uses a property of variance, $a^2 E(X) + b$.
20. A player pays $5 to play a game with a spinner. The spinner has three sectors: Red ($P=0.5$), Blue ($P=0.3$), and Green ($P=0.2$). Landing on Red wins $2, and landing on Blue wins $5. What must be the prize for landing on Green for the game to be considered fair?
    Answer: $12.50 A fair game means the expected profit is zero, or the expected payout is equal to the cost of playing. The cost is $5. Let $G$ be the prize for landing on Green. The expected payout is $E(\text{Payout}) = (2)(0.5) + (5)(0.3) + (G)(0.2)$. For a fair game, $E(\text{Payout}) = 5$. $$1.0 + 1.5 + 0.2G = 5$$ $$2.5 + 0.2G = 5$$ $$0.2G = 2.5$$ $$G = \frac{2.5}{0.2} = 12.5$$ The prize for Green must be $12.50. Distractor C ($25.00) incorrectly assumes the expected payout from Green alone must equal the cost ($0.2G=5$). Distractor D ($37.50) results from setting the expected profit to $5 instead of $0.

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