IB Mathematics: Analysis and Approaches Quiz: Discrete Random Variables

20 questions · exam conditions

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Discrete Random VariablesQuestion 1 of 20

A game of chance involves paying $5 to play. A player can win a prize of $10, $5, or $k. The probability of winning $10 is 0.1, the probability of winning $5 is 0.4, and the probability of winning $k is 0.5.

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).

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Discrete Random Variables

Practice Discrete Random Variables in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Discrete Random Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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All questions

Question 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).

$1
$2
$4 (correct answer)
$5

Explanation: 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.

Question 2

A spinner has three sectors with scores 0, 2, and k. The probability of landing on the sector with score 0 is 0.5, on score 2 is 0.3, and on score k is 0.2.

The expected score from a single spin is $E(X) = 2$ . Find the variance of the score, $Var(X)$ .

4
7 (correct answer)
11
15

Explanation: 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)

Question 3

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 $E(Y)$ .

0.01
0.1
0.5 (correct answer)
1

Explanation: 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 $E(X) = (-1)(0.2) + 0(0.5) + 1(0.3) = 0.1$ .

Question 4

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 $a$ .

0.3
0.4 (correct answer)
0.6
0.7

Explanation: 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.

Question 5

A discrete random variable $X$ has the probability distribution $P(X=k) = c(k+1)$ for $k = 0, 1, 2$ . Find the variance of $X$ .

$\frac{5}{9}$ (correct answer)
$\frac{2}{3}$
$\frac{4}{3}$
$\frac{7}{3}$

Explanation: 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}

Question 6

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).

$1
$2
$4 (correct answer)
$5

Explanation: 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.

Question 7

A discrete random variable $X$ has an expected value $E(X) = 3$ . Find the value of $E(5 - 2X)$ .

-1 (correct answer)
1
2
11

Explanation: 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)

Question 8

A discrete random variable $X$ has a variance of $Var(X) = 5$ . Find the value of $Var(3X + 4)$ .

15
19
45 (correct answer)
49

Explanation: 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

Question 9

A discrete random variable $X$ can take values $n-1$ , $n$ , and $n+1$ with equal probability. Find the variance of $X$ .

0
$\frac{2}{3}$ (correct answer)
$n^2$
$n^2 + \frac{2}{3}$

Explanation: 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

Question 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 $X$ , $Var(X)$ .

0.75 (correct answer)
1.5
2.25
3

Explanation: 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)

Question 11

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 $E(Y) + Var(Y)$ .

0
1 (correct answer)
$\mu + \sigma^2$
$\frac{\mu}{\sigma} + \frac{1}{\sigma^2}$

Explanation: We need to find the expected value and variance of

Y

. First, find

E(Y)

. We can rewrite

Y

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

Question 12

A spinner has three sectors with scores 0, 2, and k. The probability of landing on the sector with score 0 is 0.5, on score 2 is 0.3, and on score k is 0.2.

The expected score from a single spin is $E(X) = 2$ . Find the variance of the score, $Var(X)$ .

4
7 (correct answer)
11
15

Explanation: 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)

Question 13

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 $E(Y) + Var(Y)$ .

0
1 (correct answer)
$\mu + \sigma^2$
$\frac{\mu}{\sigma} + \frac{1}{\sigma^2}$

Explanation: We need to find the expected value and variance of

Y

. First, find

E(Y)

. We can rewrite

Y

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

Question 14

Two fair six-sided dice are rolled. Let the random variable $X$ be the maximum of the two scores. Find $E(X)$ .

3.5
$\frac{91}{36}$
$\frac{161}{36}$ (correct answer)
7

Explanation: 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.

Question 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 $p$ .

$p$
$p^2$
$2p(1-p)$
$2p$ (correct answer)

Explanation: 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 $E(X)=2p$ .

Question 16

The probability distribution of a discrete random variable $X$ is given by $P(X=x) = \frac{k}{x}$ for $x=2, 3, 6$ . Find $Var(X)$ .

1
2 (correct answer)
3
11

Explanation: 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

Question 17

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 $X$ , $E(X)$ .

$\frac{1}{10}$
3 (correct answer)
$\frac{161}{36}$
10

Explanation: 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.

Question 18

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 $p$ .

$p$
$p^2$
$2p(1-p)$
$2p$ (correct answer)

Explanation: 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 $E(X)=2p$ .

Question 19

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 $E(Y)$ .

0.01
0.1
0.5 (correct answer)
1

Explanation: 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 $E(X) = (-1)(0.2) + 0(0.5) + 1(0.3) = 0.1$ .

Question 20

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 $b$ .

0.25
0.4
0.5 (correct answer)
0.75

Explanation: We are given three conditions.

Sum of probabilities: $a+b+c=1$
Expected value: $E(X) = 0(a) + 1(b) + 2(c) = 1 \implies b+2c=1$
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.