The binomial distribution Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level The binomial distribution questions. See exactly how to solve problems on binomial distribution, B(n,p), P(X=r), mean np.

binomial distributionB(n,p)P(X=r)mean npconditionsidentify n and p
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A fair coin is tossed 66 times. Let XX be the number of heads, so X∼B(6,0.5)X\sim B(6,0.5). Find P(X=3)P(X=3), giving your answer to 4 significant figures.

Worked solution

  1. Write the distribution

    X∼B(6, 0.5)X\sim B(6,\ 0.5)

    X follows a binomial distribution.

  2. Substitute n, p and r

    P(X=3)=(63)(0.5)3(0.5)3P(X=3)=\binom{6}{3}(0.5)^{3}(0.5)^{3}

    Substitute the known values into the formula.

  3. Compute the probability

    P(X=3)=0.3125P(X=3)=0.3125

    Evaluate the product.

Answer
P(X=3)=0.3125P(X=3)=0.3125
Question 2
2 markseasy
A spinner lands on red with probability 0.250.25. It is spun 88 times and XX is the number of reds, so X∼B(8,0.25)X\sim B(8,0.25). Find P(X=2)P(X=2) to 4 significant figures.

Worked solution

  1. Write the distribution

    X∼B(8, 0.25)X\sim B(8,\ 0.25)

    X follows a binomial distribution.

  2. Substitute n, p and r

    P(X=2)=(82)(0.25)2(0.75)6P(X=2)=\binom{8}{2}(0.25)^{2}(0.75)^{6}

    Substitute the known values into the formula.

  3. Compute the probability

    P(X=2)=0.3115P(X=2)=0.3115

    Evaluate the product.

Answer
P(X=2)=0.3115P(X=2)=0.3115
Question 3
2 markseasy
20%20\% of the sweets in a large jar are red. A child picks 1010 sweets at random and X∼B(10,0.2)X\sim B(10,0.2) is the number of red sweets. Find P(X=3)P(X=3) to 4 significant figures.

Worked solution

  1. Write the distribution

    X∼B(10, 0.2)X\sim B(10,\ 0.2)

    X follows a binomial distribution.

  2. Substitute n, p and r

    P(X=3)=(103)(0.2)3(0.8)7P(X=3)=\binom{10}{3}(0.2)^{3}(0.8)^{7}

    Substitute the known values into the formula.

  3. Compute the probability

    P(X=3)=0.2013P(X=3)=0.2013

    Evaluate the product.

Answer
P(X=3)=0.2013P(X=3)=0.2013
Question 4
2 markseasy
A basketball player scores each free throw with probability 0.40.4. In 1212 throws let X∼B(12,0.4)X\sim B(12,0.4) be the number scored. Find P(X=5)P(X=5) to 4 significant figures.

Worked solution

  1. Write the distribution

    X∼B(12, 0.4)X\sim B(12,\ 0.4)

    X follows a binomial distribution.

  2. Substitute n, p and r

    P(X=5)=(125)(0.4)5(0.6)7P(X=5)=\binom{12}{5}(0.4)^{5}(0.6)^{7}

    Substitute the known values into the formula.

  3. Compute the probability

    P(X=5)=0.227P(X=5)=0.227

    Evaluate the product.

Answer
P(X=5)=0.227P(X=5)=0.227
Question 5
2 markseasy
A seed germinates with probability 0.60.6. A gardener sows 55 seeds and X∼B(5,0.6)X\sim B(5,0.6) is the number that germinate. Find P(X=4)P(X=4) to 4 significant figures.

Worked solution

  1. Write the distribution

    X∼B(5, 0.6)X\sim B(5,\ 0.6)

    X follows a binomial distribution.

  2. Substitute n, p and r

    P(X=4)=(54)(0.6)4(0.4)1P(X=4)=\binom{5}{4}(0.6)^{4}(0.4)^{1}

    Substitute the known values into the formula.

  3. Compute the probability

    P(X=4)=0.2592P(X=4)=0.2592

    Evaluate the product.

Answer
P(X=4)=0.2592P(X=4)=0.2592

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