Further Maths Geometric and negative binomial Practice Questions

Free Further Maths Geometric and negative binomial practice questions with full step-by-step worked solutions. Covers geometric, pmf, trials-convention, E. Practise exam-style problems and check your method.

geometricpmftrials-conventionEVarle
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.
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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 spinner is spun repeatedly. On each spin the probability of a red sector is 12\frac{1}{2}. The random variable XX is the number of spins up to and including the third red sector, so that X∼NB(3,12)X\sim\text{NB}\left(3,\frac{1}{2}\right). Which of the following is the value of E(X)E(X)?
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Worked solution

  1. State the distribution and the convention it uses

    X∼NB(3,12),x=3,4,5,…X\sim\text{NB}\left(3,\frac{1}{2}\right),\quad x=3,4,5,\ldots

    XX counts the trials up to and including the 33th success, so the smallest value XX can take is 33.

  2. Write down the probability function with r=3r=3, p=12p=\frac{1}{2} and 1−p=121-p=\frac{1}{2}

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

    The xxth trial is the 33th success, and exactly 22 of the first x−1x-1 trials are successes.

  3. Quote the mean of a negative binomial distribution

    E(X)=rp=312=6E(X)=\frac{r}{p}=\frac{3}{\frac{1}{2}}=6

    The wait to the 33th success is 33 independent geometric waits, so the mean is 33 times 1/p1/p.

  4. State the final answer

    E(X)=312=6E(X)=\frac{3}{\frac{1}{2}}=6

    This is the exact value requested.

Answer
E(X)=6E(X)=6
Question 3
4 marksintermediate
A marksman fires at a distant target. On each shot the probability of a hit is 12\frac{1}{2}. The random variable XX is the number of shots up to and including the second hit, so that X∼NB(2,12)X\sim\text{NB}\left(2,\frac{1}{2}\right). Which of the following is the value of Var(X)\text{Var}(X)?
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Worked solution

  1. State the distribution and the convention it uses

    X∼NB(2,12),x=2,3,4,…X\sim\text{NB}\left(2,\frac{1}{2}\right),\quad x=2,3,4,\ldots

    XX counts the trials up to and including the 22th success, so the smallest value XX can take is 22.

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

    P(X=x)=(x−11)(12)2(12)x−2P(X=x)=\binom{x-1}{1}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{x-2}

    The xxth trial is the 22th success, and exactly 11 of the first x−1x-1 trials are successes.

  3. Quote the mean of a negative binomial distribution

    E(X)=rp=212=4E(X)=\frac{r}{p}=\frac{2}{\frac{1}{2}}=4

    The wait to the 22th success is 22 independent geometric waits, so the mean is 22 times 1/p1/p.

  4. Quote the variance of a negative binomial distribution

    Var(X)=r(1−p)p2=2×12(12)2=4\text{Var}(X)=\frac{r\left(1-p\right)}{p^{2}}=\frac{2\times\frac{1}{2}}{\left(\frac{1}{2}\right)^{2}}=4

    The variances of the 22 independent geometric waits add.

  5. Note that the variance follows from the second moment

    Var(X)=E(X2)−[E(X)]2=20−(4)2=4\text{Var}(X)=E(X^{2})-\left[E(X)\right]^{2}=20-\left(4\right)^{2}=4

    A useful check on the quoted formula.

  6. Check by direct summation: partial sum up to x=2x=2

    ∑x=22x2P(X=x)=1≈1.00000\sum_{x=2}^{2}x^{2}P(X=x)=1\approx 1.00000

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=20E(X^{2})=\text{Var}(X)+[E(X)]^{2}=20, confirming the quoted variance (limit ≈20.00000\approx 20.00000).

  7. State the final answer

    Var(X)=2×12(12)2=4\text{Var}(X)=\frac{2\times\frac{1}{2}}{\left(\frac{1}{2}\right)^{2}}=4

    This is the exact value requested.

Answer
Var(X)=4\text{Var}(X)=4
Question 4
6 markshard
Components coming off a production line are tested one at a time. On each test the probability of a faulty component is 25\frac{2}{5}. The random variable XX is the number of tests up to and including the fourth faulty component, so that X∼NB(4,25)X\sim\text{NB}\left(4,\frac{2}{5}\right). Which of the following is the value of Var(X)\text{Var}(X)?
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Worked solution

  1. State the distribution and the convention it uses

    X∼NB(4,25),x=4,5,6,…X\sim\text{NB}\left(4,\frac{2}{5}\right),\quad x=4,5,6,\ldots

    XX counts the trials up to and including the 44th success, so the smallest value XX can take is 44.

  2. Write down the probability function with r=4r=4, p=25p=\frac{2}{5} and 1−p=351-p=\frac{3}{5}

    P(X=x)=(x−13)(25)4(35)x−4P(X=x)=\binom{x-1}{3}\left(\frac{2}{5}\right)^{4}\left(\frac{3}{5}\right)^{x-4}

    The xxth trial is the 44th success, and exactly 33 of the first x−1x-1 trials are successes.

  3. Quote the mean of a negative binomial distribution

    E(X)=rp=425=10E(X)=\frac{r}{p}=\frac{4}{\frac{2}{5}}=10

    The wait to the 44th success is 44 independent geometric waits, so the mean is 44 times 1/p1/p.

  4. Quote the variance of a negative binomial distribution

    Var(X)=r(1−p)p2=4×35(25)2=15\text{Var}(X)=\frac{r\left(1-p\right)}{p^{2}}=\frac{4\times\frac{3}{5}}{\left(\frac{2}{5}\right)^{2}}=15

    The variances of the 44 independent geometric waits add.

  5. Note that the variance follows from the second moment

    Var(X)=E(X2)−[E(X)]2=115−(10)2=15\text{Var}(X)=E(X^{2})-\left[E(X)\right]^{2}=115-\left(10\right)^{2}=15

    A useful check on the quoted formula.

  6. Check by direct summation: partial sum up to x=4x=4

    ∑x=44x2P(X=x)=256625≈0.40960\sum_{x=4}^{4}x^{2}P(X=x)=\frac{256}{625}\approx 0.40960

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=115E(X^{2})=\text{Var}(X)+[E(X)]^{2}=115, confirming the quoted variance (limit ≈115.00000\approx 115.00000).

  7. Check by direct summation: partial sum up to x=5x=5

    ∑x=45x2P(X=x)=1216625≈1.94560\sum_{x=4}^{5}x^{2}P(X=x)=\frac{1216}{625}\approx 1.94560

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=115E(X^{2})=\text{Var}(X)+[E(X)]^{2}=115, confirming the quoted variance (limit ≈115.00000\approx 115.00000).

  8. Check by direct summation: partial sum up to x=6x=6

    ∑x=46x2P(X=x)=164483125≈5.26336\sum_{x=4}^{6}x^{2}P(X=x)=\frac{16448}{3125}\approx 5.26336

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=115E(X^{2})=\text{Var}(X)+[E(X)]^{2}=115, confirming the quoted variance (limit ≈115.00000\approx 115.00000).

  9. Check by direct summation: partial sum up to x=7x=7

    ∑x=47x2P(X=x)=16691215625≈10.68237\sum_{x=4}^{7}x^{2}P(X=x)=\frac{166912}{15625}\approx 10.68237

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=115E(X^{2})=\text{Var}(X)+[E(X)]^{2}=115, confirming the quoted variance (limit ≈115.00000\approx 115.00000).

  10. Check by direct summation: partial sum up to x=8x=8

    ∑x=48x2P(X=x)=141516878125≈18.11415\sum_{x=4}^{8}x^{2}P(X=x)=\frac{1415168}{78125}\approx 18.11415

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=115E(X^{2})=\text{Var}(X)+[E(X)]^{2}=115, confirming the quoted variance (limit ≈115.00000\approx 115.00000).

  11. Check by direct summation: partial sum up to x=9x=9

    ∑x=49x2P(X=x)=530151681953125≈27.14377\sum_{x=4}^{9}x^{2}P(X=x)=\frac{53015168}{1953125}\approx 27.14377

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=115E(X^{2})=\text{Var}(X)+[E(X)]^{2}=115, confirming the quoted variance (limit ≈115.00000\approx 115.00000).

  12. Record the probability of failing on a single trial

    1−p=1−25=351-p=1-\frac{2}{5}=\frac{3}{5}

    Every failure contributes one factor of 1−p1-p.

  13. State the final answer

    Var(X)=4×35(25)2=15\text{Var}(X)=\frac{4\times\frac{3}{5}}{\left(\frac{2}{5}\right)^{2}}=15

    This is the exact value requested.

Answer
Var(X)=15\text{Var}(X)=15
Question 5
9 markschallenging
A quality control officer inspects light bulbs one by one. On each inspection the probability of a defective bulb is 13\frac{1}{3}. The random variable XX is the number of inspections up to and including the second defective bulb, so that X∼NB(2,13)X\sim\text{NB}\left(2,\frac{1}{3}\right). Which of the following is the value of E(X2)E(X^{2})?
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Worked solution

  1. State the distribution and the convention it uses

    X∼NB(2,13),x=2,3,4,…X\sim\text{NB}\left(2,\frac{1}{3}\right),\quad x=2,3,4,\ldots

    XX counts the trials up to and including the 22th success, so the smallest value XX can take is 22.

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

    P(X=x)=(x−11)(13)2(23)x−2P(X=x)=\binom{x-1}{1}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{x-2}

    The xxth trial is the 22th success, and exactly 11 of the first x−1x-1 trials are successes.

  3. Quote the mean of a negative binomial distribution

    E(X)=rp=213=6E(X)=\frac{r}{p}=\frac{2}{\frac{1}{3}}=6

    The wait to the 22th success is 22 independent geometric waits, so the mean is 22 times 1/p1/p.

  4. Quote the variance of a negative binomial distribution

    Var(X)=r(1−p)p2=2×23(13)2=12\text{Var}(X)=\frac{r\left(1-p\right)}{p^{2}}=\frac{2\times\frac{2}{3}}{\left(\frac{1}{3}\right)^{2}}=12

    The variances of the 22 independent geometric waits add.

  5. Use the relationship between the second moment, the variance and the mean

    E(X2)=Var(X)+[E(X)]2=12+(6)2=48E(X^{2})=\text{Var}(X)+\left[E(X)\right]^{2}=12+\left(6\right)^{2}=48

    Rearranging Var(X)=E(X2)−[E(X)]2\text{Var}(X)=E(X^{2})-[E(X)]^{2}.

  6. Check by direct summation: partial sum up to x=2x=2

    ∑x=22x2P(X=x)=49≈0.44444\sum_{x=2}^{2}x^{2}P(X=x)=\frac{4}{9}\approx 0.44444

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=48E(X^{2})=\text{Var}(X)+[E(X)]^{2}=48, confirming the quoted variance (limit ≈48.00000\approx 48.00000).

  7. Check by direct summation: partial sum up to x=3x=3

    ∑x=23x2P(X=x)=169≈1.77778\sum_{x=2}^{3}x^{2}P(X=x)=\frac{16}{9}\approx 1.77778

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=48E(X^{2})=\text{Var}(X)+[E(X)]^{2}=48, confirming the quoted variance (limit ≈48.00000\approx 48.00000).

  8. Check by direct summation: partial sum up to x=4x=4

    ∑x=24x2P(X=x)=11227≈4.14815\sum_{x=2}^{4}x^{2}P(X=x)=\frac{112}{27}\approx 4.14815

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=48E(X^{2})=\text{Var}(X)+[E(X)]^{2}=48, confirming the quoted variance (limit ≈48.00000\approx 48.00000).

  9. Check by direct summation: partial sum up to x=5x=5

    ∑x=25x2P(X=x)=1808243≈7.44033\sum_{x=2}^{5}x^{2}P(X=x)=\frac{1808}{243}\approx 7.44033

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=48E(X^{2})=\text{Var}(X)+[E(X)]^{2}=48, confirming the quoted variance (limit ≈48.00000\approx 48.00000).

  10. Check by direct summation: partial sum up to x=6x=6

    ∑x=26x2P(X=x)=2768243≈11.39095\sum_{x=2}^{6}x^{2}P(X=x)=\frac{2768}{243}\approx 11.39095

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=48E(X^{2})=\text{Var}(X)+[E(X)]^{2}=48, confirming the quoted variance (limit ≈48.00000\approx 48.00000).

  11. Check by direct summation: partial sum up to x=7x=7

    ∑x=27x2P(X=x)=11440729≈15.69273\sum_{x=2}^{7}x^{2}P(X=x)=\frac{11440}{729}\approx 15.69273

    The partial sums increase towards E(X2)=Var(X)+[E(X)]2=48E(X^{2})=\text{Var}(X)+[E(X)]^{2}=48, confirming the quoted variance (limit ≈48.00000\approx 48.00000).

  12. Record the probability of failing on a single trial

    1−p=1−13=231-p=1-\frac{1}{3}=\frac{2}{3}

    Every failure contributes one factor of 1−p1-p.

  13. Record the convention used throughout

    X=number of trials up to and including the required successX=\text{number of trials up to and including the required success}

    The A-Level convention counts TRIALS, not failures, so the support never includes 00.

  14. Recall the geometric probability function

    P(X=x)=(1−p)x−1p,x=1,2,3,…P(X=x)=\left(1-p\right)^{x-1}p,\quad x=1,2,3,\ldots

    There are x−1x-1 failures followed by one success on the xxth trial.

  15. State the final answer

    E(X2)=48E(X^{2})=48

    This is the exact value requested.

Answer
E(X2)=48E(X^{2})=48

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