Hard Further Maths Geometric and negative binomial Questions

Challenging, exam-style Further Maths Geometric and negative binomial questions with worked solutions. Stretch yourself on the hardest negative-binomial, le, trials-convention, gt problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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})?
Show worked solution

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
Question 2
9 markschallenging
A darts player aims repeatedly at the treble twenty. On each dart the probability of a treble twenty is 16\frac{1}{6}. The random variable XX is the number of darts up to and including the first treble twenty, so that X∼Geo(16)X\sim\text{Geo}\left(\frac{1}{6}\right). Which of the following is the value of P(X>12∣X>5)P(X>12\mid X>5)?
Show worked solution

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. Write down the definition of conditional probability

    P(X>12∣X>5)=P(X>12 and X>5)P(X>5)P(X>12\mid X>5)=\frac{P(X>12\ \text{and}\ X>5)}{P(X>5)}

    Conditioning on the event X>5X>5.

  4. Simplify the intersection

    P(X>12 and X>5)=P(X>12)P(X>12\ \text{and}\ X>5)=P(X>12)

    X>12X>12 already implies X>5X>5 because 12>512>5.

  5. Write the two tail probabilities using the geometric tail formula

    P(X>12)=(56)12,P(X>5)=(56)5P(X>12)=\left(\frac{5}{6}\right)^{12},\quad P(X>5)=\left(\frac{5}{6}\right)^{5}

    X>xX>x means the first xx trials all fail.

  6. Form the quotient and cancel

    (56)12(56)5=(56)7\frac{\left(\frac{5}{6}\right)^{12}}{\left(\frac{5}{6}\right)^{5}}=\left(\frac{5}{6}\right)^{7}

    The powers subtract: this is the memoryless property.

  7. Evaluate the remaining power

    (56)7=78125279936\left(\frac{5}{6}\right)^{7}=\frac{78125}{279936}

    The answer equals P(X>7)P(X>7): the 55 failures already seen are forgotten.

  8. Check the numerator and denominator separately

    P(X>12)=2441406252176782336,P(X>5)=31257776P(X>12)=\frac{244140625}{2176782336},\quad P(X>5)=\frac{3125}{7776}

    Their quotient reproduces the answer above.

  9. Record the probability of failing on a single trial

    1−p=1−16=561-p=1-\frac{1}{6}=\frac{5}{6}

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

  10. 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.

  11. 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.

  12. Recall the geometric cumulative probability

    P(X≤x)=1−(1−p)xP(X\le x)=1-\left(1-p\right)^{x}

    The complement of 'no success in the first xx trials'.

  13. Recall the tail probability of a geometric distribution

    P(X>x)=(1−p)xP(X>x)=\left(1-p\right)^{x}

    X>xX>x means the first xx trials were all failures.

  14. Recall the mean of a geometric distribution

    E(X)=1pE(X)=\frac{1}{p}

    For the trials convention the mean is 1/p1/p, never (1−p)/p(1-p)/p.

  15. State the final answer

    P(X>12∣X>5)=78125279936P(X>12\mid X>5)=\frac{78125}{279936}

    This is the exact value requested.

Answer
P(X>12∣X>5)=78125279936P(X>12\mid X>5)=\frac{78125}{279936}
Question 3
9 markschallenging
Cards are drawn, with replacement, from a specially printed pack. On each draw the probability of an ace is 12\frac{1}{2}. The random variable XX is the number of draws up to and including the second ace, so that X∼NB(2,12)X\sim\text{NB}\left(2,\frac{1}{2}\right). Which of the following is the least value of nn for which P(X≤n)≥0.95P(X\le n)\ge 0.95?
Show worked solution

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. Write down the condition to be satisfied

    P(X≤n)≥0.95P(X\le n)\ge 0.95

    The smallest integer nn satisfying this is required.

  4. Build up the cumulative probability term by term

    P(X≤n)=∑x=2nP(X=x)P(X\le n)=\sum_{x=2}^{n}P(X=x)

    There is no closed form, so the terms are accumulated until the condition is met.

  5. Evaluate the term P(X=2)P(X=2)

    P(X=2)=14P(X=2)=\frac{1}{4}

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

  6. Evaluate the term P(X=3)P(X=3)

    P(X=3)=14P(X=3)=\frac{1}{4}

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

  7. Evaluate the term P(X=4)P(X=4)

    P(X=4)=316P(X=4)=\frac{3}{16}

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

  8. Evaluate the term P(X=5)P(X=5)

    P(X=5)=18P(X=5)=\frac{1}{8}

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

  9. Evaluate the term P(X=6)P(X=6)

    P(X=6)=564P(X=6)=\frac{5}{64}

    Substitute x=6x=6 into the probability function.

  10. Evaluate the term P(X=7)P(X=7)

    P(X=7)=364P(X=7)=\frac{3}{64}

    Substitute x=7x=7 into the probability function.

  11. Evaluate the term P(X=8)P(X=8)

    P(X=8)=7256P(X=8)=\frac{7}{256}

    Substitute x=8x=8 into the probability function.

  12. Test n=5n=5

    P(X≤5)=1316≈0.81250 < 0.95P(X\le 5)=\frac{13}{16}\approx 0.81250\ <\ 0.95

    The condition is not yet satisfied at n=5n=5.

  13. Test n=6n=6

    P(X≤6)=5764≈0.89063 < 0.95P(X\le 6)=\frac{57}{64}\approx 0.89063\ <\ 0.95

    The condition is not yet satisfied at n=6n=6.

  14. Test n=7n=7

    P(X≤7)=1516≈0.93750 < 0.95P(X\le 7)=\frac{15}{16}\approx 0.93750\ <\ 0.95

    The condition is not yet satisfied at n=7n=7.

  15. Test n=8n=8

    P(X≤8)=247256≈0.96484 ≥ 0.95P(X\le 8)=\frac{247}{256}\approx 0.96484\ \ge\ 0.95

    The condition is satisfied at n=8n=8.

  16. Confirm that n=8n=8 is the least such value

    P(X≤7)≈0.93750,P(X≤8)≈0.96484P(X\le 7)\approx 0.93750,\quad P(X\le 8)\approx 0.96484

    The cumulative probability is increasing, so no smaller nn works.

  17. Record the probability of failing on a single trial

    1−p=1−12=121-p=1-\frac{1}{2}=\frac{1}{2}

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

  18. State the final answer

    n=8n=8

    This is the exact value requested.

Answer
n=8n=8
Question 4
9 markschallenging
Raj plays the same game of chance again and again. On each game the probability of a win is 13\frac{1}{3}. The random variable XX is the number of games up to and including the third win, so that X∼NB(3,13)X\sim\text{NB}\left(3,\frac{1}{3}\right). Which of the following is the value of P(X=7)P(X=7)?
Show worked solution

Worked solution

  1. State the distribution and the convention it uses

    X∼NB(3,13),x=3,4,5,…X\sim\text{NB}\left(3,\frac{1}{3}\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=13p=\frac{1}{3} and 1−p=231-p=\frac{2}{3}

    P(X=x)=(x−12)(13)3(23)x−3P(X=x)=\binom{x-1}{2}\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)^{x-3}

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

  3. Substitute x=7x=7 into the probability function

    P(X=7)=(62)(13)3(23)4P(X=7)=\binom{6}{2}\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)^{4}

    Choose which 22 of the first 66 trials are the earlier successes.

  4. Evaluate the binomial coefficient

    (62)=6!2! 4!=15\binom{6}{2}=\frac{6!}{2!\,4!}=15

    There are 1515 orderings of the earlier successes.

  5. Evaluate the two powers

    (13)3=127,(23)4=1681\left(\frac{1}{3}\right)^{3}=\frac{1}{27},\quad\left(\frac{2}{3}\right)^{4}=\frac{16}{81}

    33 successes and 44 failures occur in total.

  6. Multiply the three factors together

    P(X=7)=15×127×1681=80729P(X=7)=15\times\frac{1}{27}\times\frac{16}{81}=\frac{80}{729}

    This is the required probability.

  7. Check with the neighbouring probability

    P(X=8)=(23)P(X=7)P(X=8)=\left(\frac{2}{3}\right)P(X=7)

    Successive probabilities differ by one extra factor of 1−p1-p.

  8. 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.

  9. 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.

  10. 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.

  11. Recall the geometric cumulative probability

    P(X≤x)=1−(1−p)xP(X\le x)=1-\left(1-p\right)^{x}

    The complement of 'no success in the first xx trials'.

  12. Recall the tail probability of a geometric distribution

    P(X>x)=(1−p)xP(X>x)=\left(1-p\right)^{x}

    X>xX>x means the first xx trials were all failures.

  13. Recall the mean of a geometric distribution

    E(X)=1pE(X)=\frac{1}{p}

    For the trials convention the mean is 1/p1/p, never (1−p)/p(1-p)/p.

  14. Recall the variance of a geometric distribution

    Var(X)=1−pp2\text{Var}(X)=\frac{1-p}{p^{2}}

    This is the standard result quoted in the formula book.

  15. State the final answer

    P(X=7)=(62)(13)3(23)4=80729P(X=7)=\binom{6}{2}\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)^{4}=\frac{80}{729}

    This is the exact value requested.

Answer
P(X=7)=80729P(X=7)=\frac{80}{729}
Question 5
9 markschallenging
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 fifth red sector, so that X∼NB(5,12)X\sim\text{NB}\left(5,\frac{1}{2}\right). Find P(X≤12)P(X\le 12), giving your answer correct to 44 decimal places.
Show worked solution

Worked solution

  1. State the distribution and the convention it uses

    X∼NB(5,12),x=5,6,7,…X\sim\text{NB}\left(5,\frac{1}{2}\right),\quad x=5,6,7,\ldots

    XX counts the trials up to and including the 55th success, so the smallest value XX can take is 55.

  2. Write down the probability function with r=5r=5, p=12p=\frac{1}{2} and 1−p=121-p=\frac{1}{2}

    P(X=x)=(x−14)(12)5(12)x−5P(X=x)=\binom{x-1}{4}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{x-5}

    The xxth trial is the 55th success, and exactly 44 of the first x−1x-1 trials are successes.

  3. Write the cumulative probability as an explicit sum

    P(X≤12)=∑x=512(x−14)(12)5(12)x−5P(X\le 12)=\sum_{x=5}^{12}\binom{x-1}{4}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{x-5}

    There is no simple closed form, so the terms are added directly. The sum starts at x=5x=5 because X≥5X\ge 5 always.

  4. Evaluate the term P(X=5)P(X=5)

    P(X=5)=132P(X=5)=\frac{1}{32}

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

  5. Evaluate the term P(X=6)P(X=6)

    P(X=6)=564P(X=6)=\frac{5}{64}

    Substitute x=6x=6 into the probability function.

  6. Evaluate the term P(X=7)P(X=7)

    P(X=7)=15128P(X=7)=\frac{15}{128}

    Substitute x=7x=7 into the probability function.

  7. Evaluate the term P(X=8)P(X=8)

    P(X=8)=35256P(X=8)=\frac{35}{256}

    Substitute x=8x=8 into the probability function.

  8. Evaluate the term P(X=9)P(X=9)

    P(X=9)=35256P(X=9)=\frac{35}{256}

    Substitute x=9x=9 into the probability function.

  9. Evaluate the term P(X=10)P(X=10)

    P(X=10)=63512P(X=10)=\frac{63}{512}

    Substitute x=10x=10 into the probability function.

  10. Evaluate the term P(X=11)P(X=11)

    P(X=11)=1051024P(X=11)=\frac{105}{1024}

    Substitute x=11x=11 into the probability function.

  11. Evaluate the term P(X=12)P(X=12)

    P(X=12)=1652048P(X=12)=\frac{165}{2048}

    Substitute x=12x=12 into the probability function.

  12. Add the terms

    P(X≤12)=16512048P(X\le 12)=\frac{1651}{2048}

    This is the required cumulative probability.

  13. Check with the equivalent binomial statement

    P(X≤12)=∑j=512(12j)(12)j(12)12−j=16512048P(X\le 12)=\sum_{j=5}^{12}\binom{12}{j}\left(\frac{1}{2}\right)^{j}\left(\frac{1}{2}\right)^{12-j}=\frac{1651}{2048}

    X≤12X\le 12 is the same event as 'at least 55 successes in the first 1212 trials', and the two calculations agree.

  14. Record the probability of failing on a single trial

    1−p=1−12=121-p=1-\frac{1}{2}=\frac{1}{2}

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

  15. Convert the exact value to a decimal

    P(X≤12)=16512048≈0.806152P(X\le 12)=\frac{1651}{2048}\approx 0.806152

    The exact fraction is converted before rounding.

  16. State the final answer to the required accuracy

    P(X≤12)=0.8062P(X\le 12)=0.8062

    Rounded correct to 44 decimal places.

Answer
P(X≤12)=0.8062P(X\le 12)=0.8062

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