Discrete Random Variables - IB Mathematics Analysis and Approaches

1. What must be the value of the prize $k$ for the game to be fair? (A fair game is one where the expected net winnings are zero).
   Answer: $4 Let $Y$ be the random variable for the net winnings. The cost to play is $5. The possible values of $Y$ are: - Win $10: Net winning = $10 - 5 = 5$ - Win $5: Net winning = $5 - 5 = 0$ - Win $k: Net winning = $k - 5$ The probabilities are $P(Y=5)=0.1$, $P(Y=0)=0.4$, and $P(Y=k-5)=0.5$. For the game to be fair, the expected net winnings $E(Y)$ must be 0. $$E(Y) = 5(0.1) + 0(0.4) + (k-5)(0.5) = 0$$ $$0.5 + 0 + 0.5(k-5) = 0$$ $$0.5(k-5) = -0.5$$ $$k-5 = -1$$ $$k = 4$$ The value of the prize $k$ must be $4.
2. For a discrete random variable $X$, it is known that $E(X) = 1.5$ and $E(X^2) = 4.25$. Find $Var(X)$.
   Answer: 2 The formula for variance is $Var(X) = E(X^2) - [E(X)]^2$. We are given $E(X) = 1.5$ and $E(X^2) = 4.25$. Substitute these values into the formula: $$Var(X) = 4.25 - (1.5)^2$$ $$Var(X) = 4.25 - 2.25$$ $$Var(X) = 2$$ The variance of $X$ is 2. Distractor B is $[E(X)]^2$. Distractor C is from subtracting $E(X)$ instead of $[E(X)]^2$. Distractor D is from $(E(X^2)-E(X))^2$.
3. A discrete random variable $X$ has the probability distribution $P(X=k) = c(k+1)$ for $k = 0, 1, 2$. Find the variance of $X$.
   Answer: $\frac{5}{9}$ First, find the constant $c$. The sum of probabilities is 1. $$c(0+1) + c(1+1) + c(2+1) = 1$$ $$c + 2c + 3c = 1 \implies 6c = 1 \implies c = \frac{1}{6}$$ The probabilities are: $P(X=0) = \frac{1}{6}$, $P(X=1) = \frac{2}{6}$, $P(X=2) = \frac{3}{6}$. Next, find $E(X)$: $$E(X) = 0\left(\frac{1}{6}\right) + 1\left(\frac{2}{6}\right) + 2\left(\frac{3}{6}\right) = \frac{0+2+6}{6} = \frac{8}{6} = \frac{4}{3}$$ Next, find $E(X^2)$: $$E(X^2) = 0^2\left(\frac{1}{6}\right) + 1^2\left(\frac{2}{6}\right) + 2^2\left(\frac{3}{6}\right) = \frac{0+2+12}{6} = \frac{14}{6} = \frac{7}{3}$$ Finally, calculate the variance: $Var(X) = E(X^2) - [E(X)]^2$ $$Var(X) = \frac{7}{3} - \left(\frac{4}{3}\right)^2 = \frac{7}{3} - \frac{16}{9} = \frac{21}{9} - \frac{16}{9} = \frac{5}{9}$$
4. A discrete random variable $X$ has the probability distribution $P(X=0)=a$, $P(X=1)=b$, and $P(X=2)=c$. Given that $E(X)=1$ and $Var(X)=0.5$, find the value of $b$.
   Answer: 0.5 We are given three conditions. 1. Sum of probabilities: $a+b+c=1$ 2. Expected value: $E(X) = 0(a) + 1(b) + 2(c) = 1 \implies b+2c=1$ 3. Variance: $Var(X) = E(X^2) - [E(X)]^2 = 0.5$ First, let's find $E(X^2)$ from the variance formula: $$E(X^2) - (1)^2 = 0.5 \implies E(X^2) = 1.5$$ Now, write out the formula for $E(X^2)$: $$E(X^2) = 0^2(a) + 1^2(b) + 2^2(c) = 1.5 \implies b+4c=1.5$$ We now have a system of two equations with $b$ and $c$: (i) $b+2c=1$ (ii) $b+4c=1.5$ Subtracting (i) from (ii): $$(b+4c) - (b+2c) = 1.5 - 1$$ $$2c = 0.5 \implies c=0.25$$ Substitute $c=0.25$ back into (i): $$b + 2(0.25) = 1 \implies b + 0.5 = 1 \implies b=0.5$$ (We can also find $a = 1 - b - c = 1 - 0.5 - 0.25 = 0.25$). The value of $b$ is 0.5.
5. A discrete random variable $X$ has a variance of $Var(X) = 5$. Find the value of $Var(3X + 4)$.
   Answer: 45 We use the property of variance: $Var(aX + b) = a^2 Var(X)$. In this case, $a = 3$ and $b = 4$. $$Var(3X + 4) = 3^2 Var(X)$$ $$= 9 \cdot Var(X)$$ Given that $Var(X) = 5$, we substitute this value: $$= 9 \cdot 5 = 45$$ The value of $Var(3X + 4)$ is 45. Distractor A results from the incorrect calculation $3Var(X)$. Distractor B results from $3Var(X)+4$. Distractor D results from $3^2Var(X)+4$.
6. The expected score from a single spin is $E(X) = 2$. Find the variance of the score, $Var(X)$.
   Answer: 7 This is a two-step problem. First, we must find the value of $k$ using the given expected value. $$E(X) = 0(0.5) + 2(0.3) + k(0.2) = 2$$ $$0 + 0.6 + 0.2k = 2$$ $$0.2k = 1.4$$ $$k = \frac{1.4}{0.2} = 7$$ Now that we know the scores are 0, 2, and 7, we can find the variance. We need $E(X^2)$. $$E(X^2) = 0^2(0.5) + 2^2(0.3) + 7^2(0.2)$$ $$E(X^2) = 0(0.5) + 4(0.3) + 49(0.2)$$ $$E(X^2) = 0 + 1.2 + 9.8 = 11$$ Finally, we use the formula $Var(X) = E(X^2) - [E(X)]^2$. $$Var(X) = 11 - (2)^2$$ $$Var(X) = 11 - 4 = 7$$ The variance is 7. Distractor A is $[E(X)]^2$. Distractor C is $E(X^2)$.
7. The discrete random variable $X$ takes values 1, 2, and 3. The probabilities are $P(X=1)=a$, $P(X=2)=b$, and $P(X=3)=0.3$. Given that $E(X)=1.9$, find the value of $a$.
   Answer: 0.4 We are given two pieces of information. First, the sum of the probabilities must be 1: $$a + b + 0.3 = 1 \implies a + b = 0.7 \quad \text{(Eq. 1)}$$ Second, the expected value is 1.9: $$E(X) = 1(a) + 2(b) + 3(0.3) = 1.9$$ $$a + 2b + 0.9 = 1.9 \implies a + 2b = 1 \quad \text{(Eq. 2)}$$ We now have a system of two linear equations. Subtracting Eq. 1 from Eq. 2: $$(a + 2b) - (a + b) = 1 - 0.7$$ $$b = 0.3$$ Substitute $b=0.3$ back into Eq. 1: $$a + 0.3 = 0.7 \implies a = 0.4$$ The value of $a$ is 0.4.
8. Two fair six-sided dice are rolled. Let the random variable $X$ be the maximum of the two scores. Find $E(X)$.
   Answer: $\frac{161}{36}$ There are $6 \times 6 = 36$ equally likely outcomes. We need to find the probability distribution for $X$, the maximum score. - $P(X=1)$: (1,1) - 1 outcome. - $P(X=2)$: (1,2), (2,1), (2,2) - 3 outcomes. - $P(X=3)$: (1,3), (3,1), (2,3), (3,2), (3,3) - 5 outcomes. - $P(X=4)$: (4,1), (4,2), (4,3), (1,4), (2,4), (3,4), (4,4) - 7 outcomes. - $P(X=5)$: 9 outcomes. - $P(X=6)$: 11 outcomes. The pattern is $P(X=k) = \frac{2k-1}{36}$. (Check: $1+3+5+7+9+11=36$). Now, calculate $E(X)$: $$E(X) = \sum_{k=1}^{6} k \cdot P(X=k)$$ $$E(X) = 1\left(\frac{1}{36}\right) + 2\left(\frac{3}{36}\right) + 3\left(\frac{5}{36}\right) + 4\left(\frac{7}{36}\right) + 5\left(\frac{9}{36}\right) + 6\left(\frac{11}{36}\right)$$ $$E(X) = \frac{1+6+15+28+45+66}{36} = \frac{161}{36}$$ Distractor A is the expected value of a single die roll. Distractor D is the expected value of the sum of two dice.
9. A random variable $X$ has an expected value $E(X)=5$ and a standard deviation of 2. Find the value of $E(X^2)$.
   Answer: 29 The standard deviation is $\sigma = 2$, so the variance is $Var(X) = \sigma^2 = 2^2 = 4$. The formula for variance connects $E(X^2)$, $E(X)$, and $Var(X)$: $$Var(X) = E(X^2) - [E(X)]^2$$ We are given $Var(X)=4$ and $E(X)=5$. We can substitute these values and solve for $E(X^2)$. $$4 = E(X^2) - (5)^2$$ $$4 = E(X^2) - 25$$ $$E(X^2) = 4 + 25 = 29$$ The value of $E(X^2)$ is 29. Distractor B comes from calculating $25-4$.
10. A discrete random variable $X$ can take the values 0 and 2. Given that $P(X=0) = 0.25$ and $P(X=2) = 0.75$, find the variance of $X$, $Var(X)$.
    Answer: 0.75 To find the variance, we use the formula $Var(X) = E(X^2) - [E(X)]^2$. First, calculate $E(X)$: $$E(X) = (0)(0.25) + (2)(0.75) = 0 + 1.5 = 1.5$$ Next, calculate $E(X^2)$: $$E(X^2) = (0^2)(0.25) + (2^2)(0.75) = (0)(0.25) + (4)(0.75) = 0 + 3 = 3$$ Now, calculate the variance: $$Var(X) = E(X^2) - [E(X)]^2 = 3 - (1.5)^2 = 3 - 2.25 = 0.75$$ Distractor B is $E(X)$. Distractor C is $[E(X)]^2$. Distractor D is $E(X^2)$.
11. A discrete random variable $X$ has the probability distribution defined by $P(X=x) = kx$ for $x=1, 2, 3, 4$. Find the expected value of $X$, $E(X)$.
    Answer: 3 First, we must find the value of the constant $k$. The sum of probabilities for a discrete random variable must be 1. $$\sum P(X=x) = 1$$ $$P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$$ $$k(1) + k(2) + k(3) + k(4) = 1$$ $$10k = 1 \implies k = \frac{1}{10}$$ Now, we can find the expected value, $E(X)$, using the formula $E(X) = \sum x \cdot P(X=x)$. $$E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) + 4 \cdot P(X=4)$$ $$E(X) = 1\left(\frac{1}{10}\right) + 2\left(\frac{2}{10}\right) + 3\left(\frac{3}{10}\right) + 4\left(\frac{4}{10}\right)$$ $$E(X) = \frac{1 + 4 + 9 + 16}{10} = \frac{30}{10} = 3$$ The correct answer is 3.
12. Let $X$ be a random variable with mean $\mu$ and variance $\sigma^2$. A new random variable $Y$ is defined by the transformation $Y = \frac{X - \mu}{\sigma}$. Find the value of $E(Y) + Var(Y)$.
    Answer: 1 We need to find the expected value and variance of $Y$. First, find $E(Y)$. We can rewrite $Y$ as $Y = \frac{1}{\sigma}X - \frac{\mu}{\sigma}$. Using the linearity of expectation, $E(aX+b) = aE(X)+b$: $$E(Y) = E\left(\frac{1}{\sigma}X - \frac{\mu}{\sigma}\right) = \frac{1}{\sigma}E(X) - \frac{\mu}{\sigma}$$ Since $E(X)=\mu$: $$E(Y) = \frac{1}{\sigma}(\mu) - \frac{\mu}{\sigma} = \frac{\mu}{\sigma} - \frac{\mu}{\sigma} = 0$$ Next, find $Var(Y)$. Using the property $Var(aX+b) = a^2Var(X)$: $$Var(Y) = Var\left(\frac{1}{\sigma}X - \frac{\mu}{\sigma}\right) = \left(\frac{1}{\sigma}\right)^2 Var(X)$$ Since $Var(X)=\sigma^2$: $$Var(Y) = \frac{1}{\sigma^2}(\sigma^2) = 1$$ The question asks for $E(Y) + Var(Y)$, which is $0 + 1 = 1$.
13. The discrete random variable $X$ has the probability distribution $P(X=1)=a$, $P(X=2)=b$ and $P(X=4)=a$. Given that $E(X)=2.5$, find the value of $b$.
    Answer: 0 The sum of the probabilities must equal 1: $$a + b + a = 1 \implies 2a + b = 1 \quad \text{(Eq. 1)}$$ The expected value is given as 2.5: $$E(X) = 1(a) + 2(b) + 4(a) = 2.5$$ $$5a + 2b = 2.5 \quad \text{(Eq. 2)}$$ We have a system of two linear equations. From Eq. 1, we can write $b = 1 - 2a$. Substitute this into Eq. 2: $$5a + 2(1 - 2a) = 2.5$$ $$5a + 2 - 4a = 2.5$$ $$a + 2 = 2.5 \implies a = 0.5$$ Now substitute the value of $a$ back into the expression for $b$: $$b = 1 - 2(0.5) = 1 - 1 = 0$$ The value of $b$ is 0.
14. A discrete random variable $X$ takes values -1, 0, and 1 with probabilities $P(X=-1)=0.2$, $P(X=0)=0.5$, and $P(X=1)=0.3$. A new random variable is defined as $Y = X^2$. Find $E(Y)$.
    Answer: 0.5 There are two ways to solve this problem. Method 1: Find the probability distribution of $Y$. The possible values for $Y=X^2$ are $(-1)^2=1$ and $0^2=0$. - $P(Y=0) = P(X=0) = 0.5$ - $P(Y=1) = P(X=-1) + P(X=1) = 0.2 + 0.3 = 0.5$ Now, calculate $E(Y)$: $$E(Y) = 0 \cdot P(Y=0) + 1 \cdot P(Y=1) = 0(0.5) + 1(0.5) = 0.5$$ Method 2: Use the definition $E(Y) = E(X^2)$. $$E(X^2) = \sum x^2 P(X=x)$$ $$E(X^2) = (-1)^2 P(X=-1) + 0^2 P(X=0) + 1^2 P(X=1)$$ $$E(X^2) = (1)(0.2) + (0)(0.5) + (1)(0.3) = 0.2 + 0 + 0.3 = 0.5$$ So, $E(Y) = 0.5$. Distractor A, 0.01, comes from calculating $[E(X)]^2$, since $E(X) = (-1)(0.2) + 0(0.5) + 1(0.3) = 0.1$.
15. A biased coin, for which the probability of landing on heads is $p$, is tossed two times. Let $X$ be the number of heads observed. Find $E(X)$ in terms of $p$.
    Answer: $2p$ The random variable $X$ can take values 0, 1, or 2. Let H be the event of getting a head, and T be the event of getting a tail. $P(H)=p$ and $P(T)=1-p$. The probability distribution for $X$ is: - $P(X=0) = P(TT) = (1-p)(1-p) = (1-p)^2$ - $P(X=1) = P(HT \text{ or } TH) = p(1-p) + (1-p)p = 2p(1-p)$ - $P(X=2) = P(HH) = p \cdot p = p^2$ Now, we calculate the expected value $E(X)$: $$E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2)$$ $$E(X) = 0 \cdot (1-p)^2 + 1 \cdot 2p(1-p) + 2 \cdot p^2$$ $$E(X) = 2p(1-p) + 2p^2$$ $$E(X) = 2p - 2p^2 + 2p^2 = 2p$$ Alternatively, for a binomial distribution, $E(X) = np$. Here $n=2$, so $E(X)=2p$.
16. The probability distribution for a random variable $X$ is given by $P(X=x) = \frac{5-x}{10}$ for $x=1, 2, 3, 4$. Find the value of $E(3X-1)$.
    Answer: 5 First, we find the expected value of $X$, $E(X)$. The probabilities for each value of $x$ are: $P(X=1) = \frac{5-1}{10} = \frac{4}{10}$ $P(X=2) = \frac{5-2}{10} = \frac{3}{10}$ $P(X=3) = \frac{5-3}{10} = \frac{2}{10}$ $P(X=4) = \frac{5-4}{10} = \frac{1}{10}$ Now, calculate $E(X)$: $$E(X) = 1\left(\frac{4}{10}\right) + 2\left(\frac{3}{10}\right) + 3\left(\frac{2}{10}\right) + 4\left(\frac{1}{10}\right)$$ $$E(X) = \frac{4+6+6+4}{10} = \frac{20}{10} = 2$$ Using the linearity of expectation, $E(aX+b) = aE(X)+b$. $$E(3X-1) = 3E(X) - 1$$ $$E(3X-1) = 3(2) - 1 = 6 - 1 = 5$$ Distractor A is $E(X)$. Distractor C is $3E(X)$.
17. A discrete random variable $X$ can take values $n-1$, $n$, and $n+1$ with equal probability. Find the variance of $X$.
    Answer: $\frac{2}{3}$ Since there are three values with equal probability, each has a probability of $\frac{1}{3}$. First, we find the expected value, $E(X)$. $$E(X) = (n-1)\frac{1}{3} + (n)\frac{1}{3} + (n+1)\frac{1}{3}$$ $$E(X) = \frac{1}{3} [(n-1) + n + (n+1)] = \frac{1}{3}(3n) = n$$ Next, we find $E(X^2)$. $$E(X^2) = (n-1)^2\frac{1}{3} + (n)^2\frac{1}{3} + (n+1)^2\frac{1}{3}$$ $$E(X^2) = \frac{1}{3} [(n^2-2n+1) + n^2 + (n^2+2n+1)]$$ $$E(X^2) = \frac{1}{3} [3n^2 + 2]$$ Finally, we calculate the variance using $Var(X) = E(X^2) - [E(X)]^2$. $$Var(X) = \frac{1}{3}(3n^2 + 2) - n^2$$ $$Var(X) = n^2 + \frac{2}{3} - n^2 = \frac{2}{3}$$ The variance is a constant $\frac{2}{3}$, independent of $n$.
18. A discrete random variable $X$ has an expected value $E(X) = 3$. Find the value of $E(5 - 2X)$.
    Answer: -1 We use the linearity property of expectation: $E(aX + b) = aE(X) + b$. In this case, $a = -2$ and $b = 5$. $$E(5 - 2X) = E(-2X + 5)$$ $$= -2E(X) + 5$$ Given that $E(X) = 3$, we substitute this value: $$= -2(3) + 5$$ $$= -6 + 5 = -1$$ The value of $E(5 - 2X)$ is -1. Distractor B comes from calculating $2E(X)-5$. Distractor C from $5-E(X)$. Distractor D from $5+2E(X)$.
19. The probability distribution of a discrete random variable $X$ is given by $P(X=x) = \frac{k}{x}$ for $x=2, 3, 6$. Find $Var(X)$.
    Answer: 2 First, we find the value of $k$ by ensuring the sum of probabilities is 1. $$\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1$$ To add the fractions, find a common denominator, which is 6. $$\frac{3k}{6} + \frac{2k}{6} + \frac{k}{6} = 1$$ $$\frac{6k}{6} = 1 \implies k=1$$ So, the probabilities are $P(X=2)=\frac{1}{2}$, $P(X=3)=\frac{1}{3}$, and $P(X=6)=\frac{1}{6}$. Next, calculate $E(X)$. $$E(X) = 2\left(\frac{1}{2}\right) + 3\left(\frac{1}{3}\right) + 6\left(\frac{1}{6}\right) = 1 + 1 + 1 = 3$$ Next, calculate $E(X^2)$. $$E(X^2) = 2^2\left(\frac{1}{2}\right) + 3^2\left(\frac{1}{3}\right) + 6^2\left(\frac{1}{6}\right) = 4\left(\frac{1}{2}\right) + 9\left(\frac{1}{3}\right) + 36\left(\frac{1}{6}\right) = 2 + 3 + 6 = 11$$ Finally, calculate the variance. $$Var(X) = E(X^2) - [E(X)]^2 = 11 - 3^2 = 11 - 9 = 2$$

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