Geometric and negative binomial Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Geometric and negative binomial questions. See exactly how to solve problems on geometric, pmf, trials-convention, E.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Ella rolls a biased die repeatedly. On each roll the probability of a six is 12\frac{1}{2}. The random variable XX is the number of rolls up to and including the first six, so that X∼Geo(12)X\sim\text{Geo}\left(\frac{1}{2}\right). Find P(X=3)P(X=3), giving your answer as an exact fraction in its simplest form.

Worked solution

  1. State the distribution and the convention it uses

    X∼Geo(12),x=1,2,3,…X\sim\text{Geo}\left(\frac{1}{2}\right),\quad x=1,2,3,\ldots

    XX counts the trials up to and including the first success, so the smallest value XX can take is 11 (not 00).

  2. Write down the probability function with p=12p=\frac{1}{2} and 1−p=121-p=\frac{1}{2}

    P(X=x)=(12)x−1(12)P(X=x)=\left(\frac{1}{2}\right)^{x-1}\left(\frac{1}{2}\right)

    There are x−1x-1 failures and then a success on the xxth trial.

  3. Substitute x=3x=3 into the probability function

    P(X=3)=(12)2(12)P(X=3)=\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)

    There are 22 failures followed by a success on trial 33.

  4. State the final answer

    P(X=3)=(12)2(12)=18P(X=3)=\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)=\frac{1}{8}

    This is the exact value requested.

Answer
P(X=3)=18P(X=3)=\frac{1}{8}
Question 2
2 markseasy
A biased coin is tossed repeatedly. On each toss the probability of a head is 16\frac{1}{6}. The random variable XX is the number of tosses up to and including the first head, so that X∼Geo(16)X\sim\text{Geo}\left(\frac{1}{6}\right). Find P(X=4)P(X=4), giving your answer as an exact fraction in its simplest form.

Worked solution

  1. State the distribution and the convention it uses

    X∼Geo(16),x=1,2,3,…X\sim\text{Geo}\left(\frac{1}{6}\right),\quad x=1,2,3,\ldots

    XX counts the trials up to and including the first success, so the smallest value XX can take is 11 (not 00).

  2. Write down the probability function with p=16p=\frac{1}{6} and 1−p=561-p=\frac{5}{6}

    P(X=x)=(56)x−1(16)P(X=x)=\left(\frac{5}{6}\right)^{x-1}\left(\frac{1}{6}\right)

    There are x−1x-1 failures and then a success on the xxth trial.

  3. Substitute x=4x=4 into the probability function

    P(X=4)=(56)3(16)P(X=4)=\left(\frac{5}{6}\right)^{3}\left(\frac{1}{6}\right)

    There are 33 failures followed by a success on trial 44.

  4. State the final answer

    P(X=4)=(56)3(16)=1251296P(X=4)=\left(\frac{5}{6}\right)^{3}\left(\frac{1}{6}\right)=\frac{125}{1296}

    This is the exact value requested.

Answer
P(X=4)=1251296P(X=4)=\frac{125}{1296}
Question 3
2 markseasy
An archer fires arrows at a target. On each shot the probability of a bullseye is 14\frac{1}{4}. The random variable XX is the number of shots up to and including the first bullseye, so that X∼Geo(14)X\sim\text{Geo}\left(\frac{1}{4}\right). Find P(X=2)P(X=2), giving your answer as an exact fraction in its simplest form.

Worked solution

  1. State the distribution and the convention it uses

    X∼Geo(14),x=1,2,3,…X\sim\text{Geo}\left(\frac{1}{4}\right),\quad x=1,2,3,\ldots

    XX counts the trials up to and including the first success, so the smallest value XX can take is 11 (not 00).

  2. Write down the probability function with p=14p=\frac{1}{4} and 1−p=341-p=\frac{3}{4}

    P(X=x)=(34)x−1(14)P(X=x)=\left(\frac{3}{4}\right)^{x-1}\left(\frac{1}{4}\right)

    There are x−1x-1 failures and then a success on the xxth trial.

  3. Substitute x=2x=2 into the probability function

    P(X=2)=(34)1(14)P(X=2)=\left(\frac{3}{4}\right)^{1}\left(\frac{1}{4}\right)

    There are 11 failures followed by a success on trial 22.

  4. State the final answer

    P(X=2)=(34)1(14)=316P(X=2)=\left(\frac{3}{4}\right)^{1}\left(\frac{1}{4}\right)=\frac{3}{16}

    This is the exact value requested.

Answer
P(X=2)=316P(X=2)=\frac{3}{16}
Question 4
2 markseasy
A basketball player takes a long sequence of free throws. On each throw the probability of a success is 13\frac{1}{3}. The random variable XX is the number of throws up to and including the first success, so that X∼Geo(13)X\sim\text{Geo}\left(\frac{1}{3}\right). Find P(X=5)P(X=5), giving your answer as an exact fraction in its simplest form.

Worked solution

  1. State the distribution and the convention it uses

    X∼Geo(13),x=1,2,3,…X\sim\text{Geo}\left(\frac{1}{3}\right),\quad x=1,2,3,\ldots

    XX counts the trials up to and including the first success, so the smallest value XX can take is 11 (not 00).

  2. Write down the probability function with p=13p=\frac{1}{3} and 1−p=231-p=\frac{2}{3}

    P(X=x)=(23)x−1(13)P(X=x)=\left(\frac{2}{3}\right)^{x-1}\left(\frac{1}{3}\right)

    There are x−1x-1 failures and then a success on the xxth trial.

  3. Substitute x=5x=5 into the probability function

    P(X=5)=(23)4(13)P(X=5)=\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)

    There are 44 failures followed by a success on trial 55.

  4. State the final answer

    P(X=5)=(23)4(13)=16243P(X=5)=\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)=\frac{16}{243}

    This is the exact value requested.

Answer
P(X=5)=16243P(X=5)=\frac{16}{243}
Question 5
2 markseasy
Components coming off a production line are tested one at a time. On each test the probability of a faulty component is 15\frac{1}{5}. The random variable XX is the number of tests up to and including the first faulty component, so that X∼Geo(15)X\sim\text{Geo}\left(\frac{1}{5}\right). Find E(X)E(X), giving your answer in exact form.

Worked solution

  1. State the distribution and the convention it uses

    X∼Geo(15),x=1,2,3,…X\sim\text{Geo}\left(\frac{1}{5}\right),\quad x=1,2,3,\ldots

    XX counts the trials up to and including the first success, so the smallest value XX can take is 11 (not 00).

  2. Write down the probability function with p=15p=\frac{1}{5} and 1−p=451-p=\frac{4}{5}

    P(X=x)=(45)x−1(15)P(X=x)=\left(\frac{4}{5}\right)^{x-1}\left(\frac{1}{5}\right)

    There are x−1x-1 failures and then a success on the xxth trial.

  3. Quote the mean of a geometric distribution

    E(X)=1p=115=5E(X)=\frac{1}{p}=\frac{1}{\frac{1}{5}}=5

    The trials convention gives E(X)=1/pE(X)=1/p; the failures convention would give (1−p)/p(1-p)/p, which is not what XX counts here.

  4. State the final answer

    E(X)=115=5E(X)=\frac{1}{\frac{1}{5}}=5

    This is the exact value requested.

Answer
E(X)=5E(X)=5

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