IB Mathematics: Analysis and Approaches Quiz: Binomial Distribution

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Binomial DistributionQuestion 1 of 20

Let $X \sim B(n, p)$ . The mean of the distribution is 0.5. Express $P(X \ge 1)$ in terms of $n$ .

1 - (\frac{2n-1}{2n})^n

1 - (\frac{1}{2n})^n

1 - (\frac{n-1}{n})^n

1 - (\frac{1}{2})^n

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Binomial Distribution

Practice Binomial Distribution 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.

What this quiz covers

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

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let $X \sim B(n, p)$ . The mean of the distribution is 0.5. Express $P(X \ge 1)$ in terms of $n$ .

$1 - (\frac{2n-1}{2n})^n$ (correct answer)
$1 - (\frac{1}{2n})^n$
$1 - (\frac{n-1}{n})^n$
$1 - (\frac{1}{2})^n$

Explanation: When you encounter a binomial distribution problem where you need to find probabilities using given parameters, start by extracting all the information from the mean to determine the relationship between variables. Given that

X \sim B(n, p)

with mean = 0.5, you can use the fact that for a binomial distribution, the mean equals

np

. Therefore:

np = 0.5

, which gives us

p = \frac{0.5}{n} = \frac{1}{2n}

. To find

P(X \geq 1)

, use the complement rule:

P(X \geq 1) = 1 - P(X = 0)

. For a binomial distribution,

P(X = 0) = \binom{n}{0}p^0(1-p)^n = (1-p)^n

. Substituting

p = \frac{1}{2n}

P(X = 0) = \left(1 - \frac{1}{2n}\right)^n = \left(\frac{2n-1}{2n}\right)^n

Therefore:

P(X \geq 1) = 1 - \left(\frac{2n-1}{2n}\right)^n

This confirms answer A is correct. Looking at the wrong answers: B gives

1 - \left(\frac{1}{2n}\right)^n

, which would result from incorrectly using

p^n

instead of

(1-p)^n

for

P(X = 0)

. C shows

1 - \left(\frac{n-1}{n}\right)^n

, suggesting confusion about the relationship between

n

and

p

. D gives

1 - \left(\frac{1}{2}\right)^n

, which ignores the derived value of

p

entirely and assumes

p = \frac{1}{2}

. Study tip: Always use the complement rule for "at least" problems in binomial distributions—it's much simpler than calculating multiple individual probabilities. Remember that the mean constraint gives you crucial information about the parameters' relationship.

Question 2

Let $X \sim B(n, p)$ . The mean of the distribution is 0.5. Express $P(X \ge 1)$ in terms of $n$ .

$1 - (\frac{2n-1}{2n})^n$ (correct answer)
$1 - (\frac{1}{2n})^n$
$1 - (\frac{n-1}{n})^n$
$1 - (\frac{1}{2})^n$

X \sim B(n, p)

with mean = 0.5, you can use the fact that for a binomial distribution, the mean equals

np

. Therefore:

np = 0.5

, which gives us

p = \frac{0.5}{n} = \frac{1}{2n}

. To find

P(X \geq 1)

, use the complement rule:

P(X \geq 1) = 1 - P(X = 0)

. For a binomial distribution,

P(X = 0) = \binom{n}{0}p^0(1-p)^n = (1-p)^n

. Substituting

p = \frac{1}{2n}

P(X = 0) = \left(1 - \frac{1}{2n}\right)^n = \left(\frac{2n-1}{2n}\right)^n

Therefore:

P(X \geq 1) = 1 - \left(\frac{2n-1}{2n}\right)^n

This confirms answer A is correct. Looking at the wrong answers: B gives

1 - \left(\frac{1}{2n}\right)^n

, which would result from incorrectly using

p^n

instead of

(1-p)^n

for

P(X = 0)

. C shows

1 - \left(\frac{n-1}{n}\right)^n

, suggesting confusion about the relationship between

n

and

p

. D gives

1 - \left(\frac{1}{2}\right)^n

, which ignores the derived value of

p

entirely and assumes

p = \frac{1}{2}

Question 3

Let $X \sim B(n, p)$ . Given that $P(X=0) = \frac{1}{256}$ and $P(X=1) = \frac{8}{256}$ , find the value of $n$ .

4
8 (correct answer)
16
32

Explanation: We are given two equations:\n1)

P(X=0) = (1-p)^n = \frac{1}{256}

\n2)

P(X=1) = np(1-p)^{n-1} = \frac{8}{256}

\nDivide equation (2) by equation (1):\n

\frac{np(1-p)^{n-1}}{(1-p)^n} = \frac{8/256}{1/256} = 8

which simplifies to

\frac{np}{1-p} = 8

.\nFrom this, we have

np = 8(1-p) \Rightarrow p = \frac{8}{n+8}

and

1-p = \frac{n}{n+8}

.\nSubstitute

1-p

back into equation (1):\n

(\frac{n}{n+8})^n = \frac{1}{256}

.\nWe can test the integer options for

n

. Let's try

n=8

:\n

(\frac{8}{8+8})^8 = (\frac{8}{16})^8 = (\frac{1}{2})^8 = \frac{1}{256}

.\nThis matches the given probability. Thus,

n=8

Question 4

An archer hits a target with probability $p=2/5$ . All shots are independent. If the archer takes 4 shots, what is the probability that they hit the target on exactly two of the first three shots, and also hit the target on the fourth shot?

$\frac{36}{125}$
$\frac{72}{625}$ (correct answer)
$\frac{96}{625}$
$\frac{108}{625}$

Explanation: This problem has two independent parts. First, the probability of hitting exactly two of the first three shots. Let

Y

be the number of hits in the first three shots, so

Y \sim B(3, 2/5)

. We need

P(Y=2)

.\n

P(Y=2) = \binom{3}{2} (2/5)^2 (3/5)^1 = 3 \cdot \frac{4}{25} \cdot \frac{3}{5} = \frac{36}{125}

.\nSecond, the probability of hitting the target on the fourth shot is simply

p=2/5

.\nSince the shots are independent, we multiply the probabilities of these two events:\nTotal Probability =

P(Y=2) \times P(\text{hit on 4th}) = \frac{36}{125} \times \frac{2}{5} = \frac{72}{625}

Question 5

Let the random variable $X$ follow a binomial distribution, $X \sim B(n, p)$ . The mean of $X$ is 6 and the variance is 4. Find the value of $P(X=5)$ .

$\binom{18}{5} (\frac{1}{3})^5 (\frac{2}{3})^{13}$ (correct answer)
$\binom{18}{5} (\frac{2}{3})^5 (\frac{1}{3})^{13}$
$\binom{6}{5} (\frac{1}{3})^5 (\frac{2}{3})^1$
$(\frac{1}{3})^5 (\frac{2}{3})^{13}$

Explanation: The mean of a binomial distribution is given by

E(X) = np

and the variance by

Var(X) = np(1-p)

.\nGiven

np=6

and

np(1-p)=4

, we can substitute the first equation into the second:\n

6(1-p) = 4 \Rightarrow 1-p = 4/6 = 2/3 \Rightarrow p = 1/3

.\nNow, we find

n

:\n

n(1/3) = 6 \Rightarrow n = 18

.\nSo,

X \sim B(18, 1/3)

.\nWe need to find

P(X=5)

:\n

P(X=5) = \binom{18}{5} (1/3)^5 (2/3)^{18-5} = \binom{18}{5} (\frac{1}{3})^5 (\frac{2}{3})^{13}

Question 6

A new drug is effective with a probability of 0.8. The drug is administered to a sequence of patients. What is the minimum number of patients that must be treated for the probability of at least one successful treatment to be greater than 0.999?

2
3
4
5 (correct answer)

Explanation: Let

X

be the number of successful treatments among

n

patients. Then

X \sim B(n, 0.8)

. We want to find the smallest integer

n

such that

P(X \ge 1) > 0.999

.\nThis is equivalent to

1 - P(X=0) > 0.999

, which simplifies to

P(X=0) < 0.001

.\nThe probability of zero successes is

P(X=0) = \binom{n}{0} (0.8)^0 (0.2)^n = (0.2)^n

.\nWe need to solve

(0.2)^n < 0.001

.\nWe can test values of

n

:\nFor

n=1

(0.2)^1 = 0.2

(not < 0.001)\nFor

n=2

(0.2)^2 = 0.04

(not < 0.001)\nFor

n=3

(0.2)^3 = 0.008

(not < 0.001)\nFor

n=4

(0.2)^4 = 0.0016

(not < 0.001)\nFor

n=5

(0.2)^5 = 0.00032

(which is < 0.001)\nTherefore, the minimum number of patients is 5.

Question 7

A fair six-sided die is rolled 4 times. Let $X$ be the number of times a '6' is rolled. Given that at least one '6' is rolled, what is the probability that exactly two '6's are rolled?

$\frac{150}{671}$ (correct answer)
$\frac{25}{216}$
$\frac{150}{1296}$
$\frac{150}{500}$

Explanation: This is a conditional probability problem involving a binomial distribution. When you see "given that" in a probability question, you need to use the conditional probability formula: P(A|B) = P(A and B) / P(B). Let's define our events clearly. You want P(exactly 2 sixes | at least 1 six). Since X follows a binomial distribution with n=4 trials and p=1/6 probability of success, you can calculate the required probabilities. First, find P(exactly 2 sixes):

P(X=2) = \binom{4}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 = 6 \cdot \frac{1}{36} \cdot \frac{25}{36} = \frac{150}{1296}

Next, find P(at least 1 six). It's easier to use the complement:

P(X \geq 1) = 1 - P(X=0) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296}

Since "exactly 2 sixes" is already contained within "at least 1 six," P(exactly 2 sixes AND at least 1 six) = P(exactly 2 sixes). Therefore:

P(X=2|X \geq 1) = \frac{150/1296}{671/1296} = \frac{150}{671}

This confirms answer A is correct. Answer B gives the unconditional probability but with wrong calculations. Answer C is just P(X=2) without applying the conditional probability formula. Answer D uses an incorrect denominator, possibly confusing this with a different probability calculation. Remember: conditional probability problems require you to restrict your sample space to the given condition, which always changes the denominator of your final calculation.

Question 8

Let $X \sim B(n, p)$ . Given that $P(X=0) = \frac{1}{256}$ and $P(X=1) = \frac{8}{256}$ , find the value of $n$ .

4
8 (correct answer)
16
32

Explanation: We are given two equations:\n1)

P(X=0) = (1-p)^n = \frac{1}{256}

\n2)

P(X=1) = np(1-p)^{n-1} = \frac{8}{256}

\nDivide equation (2) by equation (1):\n

\frac{np(1-p)^{n-1}}{(1-p)^n} = \frac{8/256}{1/256} = 8

which simplifies to

\frac{np}{1-p} = 8

.\nFrom this, we have

np = 8(1-p) \Rightarrow p = \frac{8}{n+8}

and

1-p = \frac{n}{n+8}

.\nSubstitute

1-p

back into equation (1):\n

(\frac{n}{n+8})^n = \frac{1}{256}

.\nWe can test the integer options for

n

. Let's try

n=8

:\n

(\frac{8}{8+8})^8 = (\frac{8}{16})^8 = (\frac{1}{2})^8 = \frac{1}{256}

.\nThis matches the given probability. Thus,

n=8

Question 9

The number of successful outcomes in a sequence of trials is modelled by $X \sim B(25, p)$ . The variance of $X$ is 6. Given that the probability of success is less than the probability of failure, find the mean of $X$ .

6
10 (correct answer)
15
$\sqrt{6}$

Explanation: The variance is

Var(X) = np(1-p) = 6

. With

n=25

, we have

25p(1-p) = 6

.\nThis leads to the quadratic equation

25p^2 - 25p + 6 = 0

.\nFactoring gives

(5p-2)(5p-3) = 0

, so the possible values for

p

are

p=2/5

and

p=3/5

.\nThe condition 'probability of success is less than the probability of failure' means

p < 1-p

, which simplifies to

2p < 1

p < 0.5

.\nWe must choose

p=2/5 = 0.4

.\nThe mean is

E(X) = np = 25 \cdot (2/5) = 10

Question 10

6
10 (correct answer)
15
$\sqrt{6}$

Explanation: The variance is

Var(X) = np(1-p) = 6

. With

n=25

, we have

25p(1-p) = 6

.\nThis leads to the quadratic equation

25p^2 - 25p + 6 = 0

.\nFactoring gives

(5p-2)(5p-3) = 0

, so the possible values for

p

are

p=2/5

and

p=3/5

.\nThe condition 'probability of success is less than the probability of failure' means

p < 1-p

, which simplifies to

2p < 1

p < 0.5

.\nWe must choose

p=2/5 = 0.4

.\nThe mean is

E(X) = np = 25 \cdot (2/5) = 10

Question 11

For a random variable $X \sim B(15, 2/5)$ , for which integer value of $k$ is the probability $P(X=k)$ at its maximum?

5
6 (correct answer)
7
8

Explanation: The value of

k

that maximizes

P(X=k)

is the mode of the distribution. For a binomial distribution

B(n, p)

, the mode is given by

\lfloor (n+1)p \rfloor

.\nHere,

n=15

and

p=2/5=0.4

.\nWe calculate

(n+1)p = (15+1)(2/5) = 16 \cdot \frac{2}{5} = \frac{32}{5} = 6.4

.\nSince 6.4 is not an integer, the mode is unique and is equal to

\lfloor 6.4 \rfloor = 6

.\nThe mean is

np = 15(2/5) = 6

. The mode is always the integer(s) closest to the mean, which in this case is 6.

Question 12

$\frac{36}{125}$
$\frac{72}{625}$ (correct answer)
$\frac{96}{625}$
$\frac{108}{625}$

Explanation: This problem has two independent parts. First, the probability of hitting exactly two of the first three shots. Let

Y

be the number of hits in the first three shots, so

Y \sim B(3, 2/5)

. We need

P(Y=2)

.\n

P(Y=2) = \binom{3}{2} (2/5)^2 (3/5)^1 = 3 \cdot \frac{4}{25} \cdot \frac{3}{5} = \frac{36}{125}

.\nSecond, the probability of hitting the target on the fourth shot is simply

p=2/5

.\nSince the shots are independent, we multiply the probabilities of these two events:\nTotal Probability =

P(Y=2) \times P(\text{hit on 4th}) = \frac{36}{125} \times \frac{2}{5} = \frac{72}{625}

Question 13

2
3
4
5 (correct answer)

Explanation: Let

X

be the number of successful treatments among

n

patients. Then

X \sim B(n, 0.8)

. We want to find the smallest integer

n

such that

P(X \ge 1) > 0.999

.\nThis is equivalent to

1 - P(X=0) > 0.999

, which simplifies to

P(X=0) < 0.001

.\nThe probability of zero successes is

P(X=0) = \binom{n}{0} (0.8)^0 (0.2)^n = (0.2)^n

.\nWe need to solve

(0.2)^n < 0.001

.\nWe can test values of

n

:\nFor

n=1

(0.2)^1 = 0.2

(not < 0.001)\nFor

n=2

(0.2)^2 = 0.04

(not < 0.001)\nFor

n=3

(0.2)^3 = 0.008

(not < 0.001)\nFor

n=4

(0.2)^4 = 0.0016

(not < 0.001)\nFor

n=5

(0.2)^5 = 0.00032

(which is < 0.001)\nTherefore, the minimum number of patients is 5.

Question 14

Let $X \sim B(4, 1/3)$ . Find the probability $P(1 \le X < 3)$ .

$\frac{8}{27}$
$\frac{16}{27}$
$\frac{56}{81}$ (correct answer)
$\frac{64}{81}$

Explanation: We need to find

P(1 \le X < 3)

, which is equal to

P(X=1) + P(X=2)

.\nFirst, calculate

P(X=1)

:\n

P(X=1) = \binom{4}{1} (1/3)^1 (2/3)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81}

.\nNext, calculate

P(X=2)

:\n

P(X=2) = \binom{4}{2} (1/3)^2 (2/3)^2 = 6 \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81}

.\nNow, add the probabilities:\n

P(1 \le X < 3) = \frac{32}{81} + \frac{24}{81} = \frac{56}{81}

Question 15

For a random variable $X \sim B(n, 1/4)$ , it is found that $P(X=2) = \frac{9}{2} \times P(X=1)$ . Find the value of $n$ .

10
18
19
28 (correct answer)

Explanation: We are given the relation

P(X=2) = \frac{9}{2} P(X=1)

. Let's write this out using the binomial formula:\n

\binom{n}{2} (1/4)^2 (3/4)^{n-2} = \frac{9}{2} \cdot \binom{n}{1} (1/4)^1 (3/4)^{n-1}

.\n

\frac{n(n-1)}{2} \cdot \frac{1}{16} \cdot (3/4)^{n-2} = \frac{9}{2} \cdot n \cdot \frac{1}{4} \cdot (3/4)^{n-1}

.\nAssuming

n \ge 2

, we can divide by

n

and other common terms. To simplify, divide both sides by

(1/4)(3/4)^{n-2}

:\n

\frac{n-1}{2} \cdot \frac{1}{4} = \frac{9}{2} \cdot (3/4)

.\n

\frac{n-1}{8} = \frac{27}{8}

.\n

n-1 = 27

, so

n = 28

Question 16

For a random variable $X \sim B(15, 2/5)$ , for which integer value of $k$ is the probability $P(X=k)$ at its maximum?

5
6 (correct answer)
7
8

Explanation: The value of

k

that maximizes

P(X=k)

is the mode of the distribution. For a binomial distribution

B(n, p)

, the mode is given by

\lfloor (n+1)p \rfloor

.\nHere,

n=15

and

p=2/5=0.4

.\nWe calculate

(n+1)p = (15+1)(2/5) = 16 \cdot \frac{2}{5} = \frac{32}{5} = 6.4

.\nSince 6.4 is not an integer, the mode is unique and is equal to

\lfloor 6.4 \rfloor = 6

.\nThe mean is

np = 15(2/5) = 6

. The mode is always the integer(s) closest to the mean, which in this case is 6.

Question 17

Let $X \sim B(4, p)$ . Given that the mean of $X$ is 1, find the probability that the number of successes is an extreme outcome (either 0 or 4).

$\frac{1}{256}$
$\frac{81}{256}$
$\frac{82}{256}$ (correct answer)
$\frac{108}{256}$

Explanation: First, find the value of

p

. The mean is

E(X) = np = 1

.\nGiven

n=4

, we have

4p = 1

, so

p = 1/4

.\nWe need to find

P(X=0) + P(X=4)

.\n

P(X=0) = \binom{4}{0} (1/4)^0 (3/4)^4 = (3/4)^4 = \frac{81}{256}

.\n

P(X=4) = \binom{4}{4} (1/4)^4 (3/4)^0 = (1/4)^4 = \frac{1}{256}

.\nThe sum is

P(X=0) + P(X=4) = \frac{81}{256} + \frac{1}{256} = \frac{82}{256}

Question 18

For a random variable $X \sim B(4, 1/3)$ , find the value of $P(X=2 | X \ge 1)$ .

$\frac{8}{65}$
$\frac{24}{65}$ (correct answer)
$\frac{24}{49}$
$\frac{8}{27}$

Explanation: The conditional probability is given by

P(X=2 | X \ge 1) = \frac{P(X=2 \text{ and } X \ge 1)}{P(X \ge 1)}

.\nSince the event

X=2

is a subset of the event

X \ge 1

, their intersection is just

X=2

. So we need to calculate

\frac{P(X=2)}{P(X \ge 1)}

.\nFirst,

P(X \ge 1) = 1 - P(X=0)

.\n

P(X=0) = \binom{4}{0} (1/3)^0 (2/3)^4 = (2/3)^4 = 16/81

.\nSo,

P(X \ge 1) = 1 - 16/81 = 65/81

.\nNext,

P(X=2) = \binom{4}{2} (1/3)^2 (2/3)^2 = 6 \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81}

.\nTherefore,

P(X=2 | X \ge 1) = \frac{24/81}{65/81} = \frac{24}{65}

Question 19

For a random variable $X \sim B(n, 1/4)$ , it is found that $P(X=2) = \frac{9}{2} \times P(X=1)$ . Find the value of $n$ .

10
18
19
28 (correct answer)

Explanation: We are given the relation

P(X=2) = \frac{9}{2} P(X=1)

. Let's write this out using the binomial formula:\n

\binom{n}{2} (1/4)^2 (3/4)^{n-2} = \frac{9}{2} \cdot \binom{n}{1} (1/4)^1 (3/4)^{n-1}

.\n

\frac{n(n-1)}{2} \cdot \frac{1}{16} \cdot (3/4)^{n-2} = \frac{9}{2} \cdot n \cdot \frac{1}{4} \cdot (3/4)^{n-1}

.\nAssuming

n \ge 2

, we can divide by

n

and other common terms. To simplify, divide both sides by

(1/4)(3/4)^{n-2}

:\n

\frac{n-1}{2} \cdot \frac{1}{4} = \frac{9}{2} \cdot (3/4)

.\n

\frac{n-1}{8} = \frac{27}{8}

.\n

n-1 = 27

, so

n = 28

Question 20

In which of the following scenarios can the random variable $X$ be correctly modelled by a binomial distribution?

A student takes a 15-question multiple-choice test, where each question has 4 options. The student guesses every answer. $X$ is the number of correct answers. (correct answer)
A quality inspector checks 20 items from a small batch of 100, which contains 10 defective items. $X$ is the number of defective items found by the inspector.
A person plays a game where they roll a fair six-sided die. If it shows a 6, they win. They play until they have won 3 times. $X$ is the total number of rolls.
The number of rainy days in a specific week in London. $X$ is the number of days it rains.

Explanation: A binomial distribution requires four conditions: 1. A fixed number of trials (

n

). 2. Each trial must be independent. 3. Each trial has only two outcomes (success/failure). 4. The probability of success (

p

) is constant for each trial.\nA: This fits the model perfectly.

n=15

is fixed, guessing is independent, each question is either correct (success) or incorrect (failure), and

p=1/4

is constant. This is a binomial distribution.\nB: This is sampling without replacement from a small population. The probability of drawing a defective item changes after each draw, so the trials are not independent. This is a hypergeometric distribution.\nC: The number of trials is not fixed; it continues until a certain number of successes are achieved. This describes a negative binomial distribution.\nD: The weather on one day is often dependent on the weather of the previous day, so the independence assumption is likely violated.