Hard Further Maths Discrete random variables Questions

Challenging, exam-style Further Maths Discrete random variables questions with worked solutions. Stretch yourself on the hardest discrete-random-variables, variance, linear-transformation, expectation problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The discrete random variable XX has probability distribution given by P(X=x)=x10P(X=x)=\frac{x}{10} for x=1,2,3,4x=1,2,3,4. Which of the following is the exact value of Var(3X−1)\text{Var}(3X-1)?
Show worked solution

Worked solution

  1. Evaluate the probability function at each value of xx

    P(X=1)=110,P(X=2)=15,P(X=3)=310,P(X=4)=25P(X=1)=\frac{1}{10},\quad P(X=2)=\frac{1}{5},\quad P(X=3)=\frac{3}{10},\quad P(X=4)=\frac{2}{5}

    Substituting each value of xx into the formula gives its probability.

  2. Write out the completed probability distribution

    x1234P(X=x)1101531025\begin{array}{c|cccc}x&1&2&3&4\\\hline P(X=x)&\frac{1}{10}&\frac{1}{5}&\frac{3}{10}&\frac{2}{5}\end{array}

    With kk known, every probability is a definite fraction.

  3. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  4. Multiply each value of xx by its probability

    E(X)=1×110+2×15+3×310+4×25E(X)=1\times\frac{1}{10}+2\times\frac{1}{5}+3\times\frac{3}{10}+4\times\frac{2}{5}

    This is the sum defining E(X)E(X).

  5. Add the contributions to E(X)E(X)

    110+25+910+85=3\frac{1}{10}+\frac{2}{5}+\frac{9}{10}+\frac{8}{5}=3

    Adding the exact fractions gives the required expectation.

  6. Recall the definition of E(X2)E(X^{2})

    E(X2)=∑xx2P(X=x)E(X^{2})=\sum_{x}x^{2}P(X=x)

    Each value of x2x^{2} is multiplied by the probability of that value of xx.

  7. Evaluate x2x^{2} at each value of xx

    x=1⇒x2=1,x=2⇒x2=4,x=3⇒x2=9,x=4⇒x2=16x=1\Rightarrow x^{2}=1,\quad x=2\Rightarrow x^{2}=4,\quad x=3\Rightarrow x^{2}=9,\quad x=4\Rightarrow x^{2}=16

    Apply the function to each value of xx before weighting.

  8. Multiply each value of x2x^{2} by its probability

    E(X2)=1×110+4×15+9×310+16×25E(X^{2})=1\times\frac{1}{10}+4\times\frac{1}{5}+9\times\frac{3}{10}+16\times\frac{2}{5}

    This is the sum defining E(X2)E(X^{2}).

  9. Add the contributions to E(X2)E(X^{2})

    110+45+2710+325=10\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}=10

    Adding the exact fractions gives the required expectation.

  10. Recall the formula for Var(X)\text{Var}(X)

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

    The variance is the mean of the squares minus the square of the mean.

  11. Substitute E(X2)E(X^{2}) and E(X)E(X) into the variance formula

    Var(X)=10−(3)2\text{Var}(X)=10-\left(3\right)^{2}

    Both expectations are already known as exact fractions.

  12. Simplify to find Var(X)\text{Var}(X)

    Var(X)=1\text{Var}(X)=1

    Subtracting the exact fractions gives the variance.

  13. Recall the rule for the variance of a linear function of XX

    Var(aX+b)=a2Var(X)\text{Var}(aX+b)=a^{2}\text{Var}(X)

    The multiplier is SQUARED and the constant bb disappears entirely.

  14. Substitute Var(X)\text{Var}(X) and a2=9a^{2}=9

    Var(3X−1)=9×1\text{Var}(3X-1)=9\times1

    Here a=3a=3, so a2=9a^{2}=9.

  15. State the value of E(X)E(X)

    E(X)=3E(X)=3

    This is the exact value of E(X)E(X).

  16. Select the option equal to this value

    Var(3X−1)=9\text{Var}(3X-1)=9

    The constant term has vanished and the multiplier has been squared.

Answer
Var(3X−1)=9\text{Var}(3X-1)=9
Question 2
9 markschallenging
The discrete random variable XX has probability distribution given by P(X=x)=kxP(X=x)=kx for x=2,4,6x=2,4,6, where kk is a constant. Which of the following is the exact value of Var(X)\text{Var}(X)?
Show worked solution

Worked solution

  1. Evaluate the probability function at each value of xx

    P(X=2)=2k,P(X=4)=4k,P(X=6)=6kP(X=2)=2k,\quad P(X=4)=4k,\quad P(X=6)=6k

    Substituting each value of xx into the formula gives its probability.

  2. Use the fact that the probabilities sum to 11

    ∑xP(X=x)=1\sum_{x}P(X=x)=1

    The total probability of a discrete random variable is always 11.

  3. Add the probabilities in terms of kk

    2k+4k+6k=12k+4k+6k=1

    Every entry of the distribution is written in terms of kk.

  4. Collect like terms to form an equation in kk

    12k=112k=1

    The equation is linear in kk.

  5. Solve the equation for kk

    12k=1⇒k=11212k=1\Rightarrow k=\frac{1}{12}

    Dividing by the coefficient of kk gives the value of kk.

  6. Write out the completed probability distribution

    x246P(X=x)161312\begin{array}{c|ccc}x&2&4&6\\\hline P(X=x)&\frac{1}{6}&\frac{1}{3}&\frac{1}{2}\end{array}

    With kk known, every probability is a definite fraction.

  7. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  8. Multiply each value of xx by its probability

    E(X)=2×16+4×13+6×12E(X)=2\times\frac{1}{6}+4\times\frac{1}{3}+6\times\frac{1}{2}

    This is the sum defining E(X)E(X).

  9. Add the contributions to E(X)E(X)

    13+43+3=143\frac{1}{3}+\frac{4}{3}+3=\frac{14}{3}

    Adding the exact fractions gives the required expectation.

  10. Recall the definition of E(X2)E(X^{2})

    E(X2)=∑xx2P(X=x)E(X^{2})=\sum_{x}x^{2}P(X=x)

    Each value of x2x^{2} is multiplied by the probability of that value of xx.

  11. Evaluate x2x^{2} at each value of xx

    x=2⇒x2=4,x=4⇒x2=16,x=6⇒x2=36x=2\Rightarrow x^{2}=4,\quad x=4\Rightarrow x^{2}=16,\quad x=6\Rightarrow x^{2}=36

    Apply the function to each value of xx before weighting.

  12. Multiply each value of x2x^{2} by its probability

    E(X2)=4×16+16×13+36×12E(X^{2})=4\times\frac{1}{6}+16\times\frac{1}{3}+36\times\frac{1}{2}

    This is the sum defining E(X2)E(X^{2}).

  13. Add the contributions to E(X2)E(X^{2})

    23+163+18=24\frac{2}{3}+\frac{16}{3}+18=24

    Adding the exact fractions gives the required expectation.

  14. Recall the formula for Var(X)\text{Var}(X)

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

    The variance is the mean of the squares minus the square of the mean.

  15. Substitute E(X2)E(X^{2}) and E(X)E(X) into the variance formula

    Var(X)=24−(143)2\text{Var}(X)=24-\left(\frac{14}{3}\right)^{2}

    Both expectations are already known as exact fractions.

  16. Select the option equal to this value

    Var(X)=24−1969=209\text{Var}(X)=24-\frac{196}{9}=\frac{20}{9}

    This is the exact variance of XX.

Answer
Var(X)=24−1969=209\text{Var}(X)=24-\frac{196}{9}=\frac{20}{9}
Question 3
9 markschallenging
The discrete random variable XX has the probability distribution x−1012P(X=x)k2k2kk\begin{array}{c|cccc}x&-1&0&1&2\\\hline P(X=x)&k&2k&2k&k\end{array}, where kk is a constant. Which of the following is the exact value of Var(4X+3)\text{Var}(4X+3)?
Show worked solution

Worked solution

  1. Use the fact that the probabilities sum to 11

    ∑xP(X=x)=1\sum_{x}P(X=x)=1

    The total probability of a discrete random variable is always 11.

  2. Add the probabilities in terms of kk

    k+2k+2k+k=1k+2k+2k+k=1

    Every entry of the distribution is written in terms of kk.

  3. Collect like terms to form an equation in kk

    6k=16k=1

    The equation is linear in kk.

  4. Solve the equation for kk

    6k=1⇒k=166k=1\Rightarrow k=\frac{1}{6}

    Dividing by the coefficient of kk gives the value of kk.

  5. Write out the completed probability distribution

    x−1012P(X=x)16131316\begin{array}{c|cccc}x&-1&0&1&2\\\hline P(X=x)&\frac{1}{6}&\frac{1}{3}&\frac{1}{3}&\frac{1}{6}\end{array}

    With kk known, every probability is a definite fraction.

  6. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  7. Multiply each value of xx by its probability

    E(X)=(−1)×16+0×13+1×13+2×16E(X)=\left(-1\right)\times\frac{1}{6}+0\times\frac{1}{3}+1\times\frac{1}{3}+2\times\frac{1}{6}

    This is the sum defining E(X)E(X).

  8. Add the contributions to E(X)E(X)

    −16+0+13+13=12-\frac{1}{6}+0+\frac{1}{3}+\frac{1}{3}=\frac{1}{2}

    Adding the exact fractions gives the required expectation.

  9. Recall the definition of E(X2)E(X^{2})

    E(X2)=∑xx2P(X=x)E(X^{2})=\sum_{x}x^{2}P(X=x)

    Each value of x2x^{2} is multiplied by the probability of that value of xx.

  10. Evaluate x2x^{2} at each value of xx

    x=−1⇒x2=1,x=0⇒x2=0,x=1⇒x2=1,x=2⇒x2=4x=-1\Rightarrow x^{2}=1,\quad x=0\Rightarrow x^{2}=0,\quad x=1\Rightarrow x^{2}=1,\quad x=2\Rightarrow x^{2}=4

    Apply the function to each value of xx before weighting.

  11. Multiply each value of x2x^{2} by its probability

    E(X2)=1×16+0×13+1×13+4×16E(X^{2})=1\times\frac{1}{6}+0\times\frac{1}{3}+1\times\frac{1}{3}+4\times\frac{1}{6}

    This is the sum defining E(X2)E(X^{2}).

  12. Add the contributions to E(X2)E(X^{2})

    16+0+13+23=76\frac{1}{6}+0+\frac{1}{3}+\frac{2}{3}=\frac{7}{6}

    Adding the exact fractions gives the required expectation.

  13. Recall the formula for Var(X)\text{Var}(X)

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

    The variance is the mean of the squares minus the square of the mean.

  14. Substitute E(X2)E(X^{2}) and E(X)E(X) into the variance formula

    Var(X)=76−(12)2\text{Var}(X)=\frac{7}{6}-\left(\frac{1}{2}\right)^{2}

    Both expectations are already known as exact fractions.

  15. Simplify to find Var(X)\text{Var}(X)

    Var(X)=1112\text{Var}(X)=\frac{11}{12}

    Subtracting the exact fractions gives the variance.

  16. Recall the rule for the variance of a linear function of XX

    Var(aX+b)=a2Var(X)\text{Var}(aX+b)=a^{2}\text{Var}(X)

    The multiplier is SQUARED and the constant bb disappears entirely.

  17. Substitute Var(X)\text{Var}(X) and a2=16a^{2}=16

    Var(4X+3)=16×1112\text{Var}(4X+3)=16\times\frac{11}{12}

    Here a=4a=4, so a2=16a^{2}=16.

  18. Select the option equal to this value

    Var(4X+3)=443\text{Var}(4X+3)=\frac{44}{3}

    The constant term has vanished and the multiplier has been squared.

Answer
Var(4X+3)=443\text{Var}(4X+3)=\frac{44}{3}
Question 4
9 markschallenging
The discrete random variable XX has the probability distribution x1234P(X=x)16k132k\begin{array}{c|cccc}x&1&2&3&4\\\hline P(X=x)&\frac{1}{6}&k&\frac{1}{3}&2k\end{array}, where kk is a constant. Find the exact value of Var(X)\text{Var}(X).
Show worked solution

Worked solution

  1. Use the fact that the probabilities sum to 11

    ∑xP(X=x)=1\sum_{x}P(X=x)=1

    The total probability of a discrete random variable is always 11.

  2. Add the probabilities in terms of kk

    16+k+13+2k=1\frac{1}{6}+k+\frac{1}{3}+2k=1

    Every entry of the distribution is written in terms of kk.

  3. Collect like terms to form an equation in kk

    3k+12=13k+\frac{1}{2}=1

    The equation is linear in kk.

  4. Solve the equation for kk

    3k+12=1⇒k=163k+\frac{1}{2}=1\Rightarrow k=\frac{1}{6}

    Dividing by the coefficient of kk gives the value of kk.

  5. Write out the completed probability distribution

    x1234P(X=x)16161313\begin{array}{c|cccc}x&1&2&3&4\\\hline P(X=x)&\frac{1}{6}&\frac{1}{6}&\frac{1}{3}&\frac{1}{3}\end{array}

    With kk known, every probability is a definite fraction.

  6. Recall the definition of E(X)E(X)

    E(X)=∑xxP(X=x)E(X)=\sum_{x}xP(X=x)

    Each value of xx is multiplied by the probability of that value of xx.

  7. Multiply each value of xx by its probability

    E(X)=1×16+2×16+3×13+4×13E(X)=1\times\frac{1}{6}+2\times\frac{1}{6}+3\times\frac{1}{3}+4\times\frac{1}{3}

    This is the sum defining E(X)E(X).

  8. Add the contributions to E(X)E(X)

    16+13+1+43=176\frac{1}{6}+\frac{1}{3}+1+\frac{4}{3}=\frac{17}{6}

    Adding the exact fractions gives the required expectation.

  9. Recall the definition of E(X2)E(X^{2})

    E(X2)=∑xx2P(X=x)E(X^{2})=\sum_{x}x^{2}P(X=x)

    Each value of x2x^{2} is multiplied by the probability of that value of xx.

  10. Evaluate x2x^{2} at each value of xx

    x=1⇒x2=1,x=2⇒x2=4,x=3⇒x2=9,x=4⇒x2=16x=1\Rightarrow x^{2}=1,\quad x=2\Rightarrow x^{2}=4,\quad x=3\Rightarrow x^{2}=9,\quad x=4\Rightarrow x^{2}=16

    Apply the function to each value of xx before weighting.

  11. Multiply each value of x2x^{2} by its probability

    E(X2)=1×16+4×16+9×13+16×13E(X^{2})=1\times\frac{1}{6}+4\times\frac{1}{6}+9\times\frac{1}{3}+16\times\frac{1}{3}

    This is the sum defining E(X2)E(X^{2}).

  12. Add the contributions to E(X2)E(X^{2})

    16+23+3+163=556\frac{1}{6}+\frac{2}{3}+3+\frac{16}{3}=\frac{55}{6}

    Adding the exact fractions gives the required expectation.

  13. Recall the formula for Var(X)\text{Var}(X)

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

    The variance is the mean of the squares minus the square of the mean.

  14. Substitute E(X2)E(X^{2}) and E(X)E(X) into the variance formula

    Var(X)=556−(176)2\text{Var}(X)=\frac{55}{6}-\left(\frac{17}{6}\right)^{2}

    Both expectations are already known as exact fractions.

  15. State the value of E(X)E(X)

    E(X)=176E(X)=\frac{17}{6}

    This is the exact value of E(X)E(X).

  16. State the value of Var(X)\text{Var}(X)

    Var(X)=556−28936=4136\text{Var}(X)=\frac{55}{6}-\frac{289}{36}=\frac{41}{36}

    This is the exact variance of XX.

Answer
Var(X)=556−28936=4136\text{Var}(X)=\frac{55}{6}-\frac{289}{36}=\frac{41}{36}
Question 5
9 markschallenging
The discrete random variable XX has the probability distribution x12345P(X=x)kk2k2k4k\begin{array}{c|ccccc}x&1&2&3&4&5\\\hline P(X=x)&k&k&2k&2k&4k\end{array}, where kk is a constant. Find the exact value of E(X(X−1))E(X\left(X-1\right)).
Show worked solution

Worked solution

  1. Use the fact that the probabilities sum to 11

    ∑xP(X=x)=1\sum_{x}P(X=x)=1

    The total probability of a discrete random variable is always 11.

  2. Add the probabilities in terms of kk

    k+k+2k+2k+4k=1k+k+2k+2k+4k=1

    Every entry of the distribution is written in terms of kk.

  3. Collect like terms to form an equation in kk

    10k=110k=1

    The equation is linear in kk.

  4. Solve the equation for kk

    10k=1⇒k=11010k=1\Rightarrow k=\frac{1}{10}

    Dividing by the coefficient of kk gives the value of kk.

  5. Write out the completed probability distribution

    x12345P(X=x)110110151525\begin{array}{c|ccccc}x&1&2&3&4&5\\\hline P(X=x)&\frac{1}{10}&\frac{1}{10}&\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\end{array}

    With kk known, every probability is a definite fraction.

  6. Recall the definition of E(X(X−1))E(X\left(X-1\right))

    E(X(X−1))=∑xx(x−1)P(X=x)E(X\left(X-1\right))=\sum_{x}x\left(x-1\right)P(X=x)

    Each value of x(x−1)x\left(x-1\right) is multiplied by the probability of that value of xx.

  7. Evaluate x(x−1)x\left(x-1\right) at each value of xx

    x=1⇒x(x−1)=0,x=2⇒x(x−1)=2,x=3⇒x(x−1)=6,x=4⇒x(x−1)=12,x=5⇒x(x−1)=20x=1\Rightarrow x\left(x-1\right)=0,\quad x=2\Rightarrow x\left(x-1\right)=2,\quad x=3\Rightarrow x\left(x-1\right)=6,\quad x=4\Rightarrow x\left(x-1\right)=12,\quad x=5\Rightarrow x\left(x-1\right)=20

    Apply the function to each value of xx before weighting.

  8. Multiply each value of x(x−1)x\left(x-1\right) by its probability

    E(X(X−1))=0×110+2×110+6×15+12×15+20×25E(X\left(X-1\right))=0\times\frac{1}{10}+2\times\frac{1}{10}+6\times\frac{1}{5}+12\times\frac{1}{5}+20\times\frac{2}{5}

    This is the sum defining E(X(X−1))E(X\left(X-1\right)).

  9. Add the contributions to E(X(X−1))E(X\left(X-1\right))

    0+15+65+125+8=5950+\frac{1}{5}+\frac{6}{5}+\frac{12}{5}+8=\frac{59}{5}

    Adding the exact fractions gives the required expectation.

  10. Find the contribution to E(X(X−1))E(X\left(X-1\right)) from x=1x=1

    0×110=00\times\frac{1}{10}=0

    The contribution of a single value of xx to E(X(X−1))E(X\left(X-1\right)).

  11. Find the contribution to E(X(X−1))E(X\left(X-1\right)) from x=2x=2

    2×110=152\times\frac{1}{10}=\frac{1}{5}

    The contribution of a single value of xx to E(X(X−1))E(X\left(X-1\right)).

  12. Find the contribution to E(X(X−1))E(X\left(X-1\right)) from x=3x=3

    6×15=656\times\frac{1}{5}=\frac{6}{5}

    The contribution of a single value of xx to E(X(X−1))E(X\left(X-1\right)).

  13. Find the contribution to E(X(X−1))E(X\left(X-1\right)) from x=4x=4

    12×15=12512\times\frac{1}{5}=\frac{12}{5}

    The contribution of a single value of xx to E(X(X−1))E(X\left(X-1\right)).

  14. Find the contribution to E(X(X−1))E(X\left(X-1\right)) from x=5x=5

    20×25=820\times\frac{2}{5}=8

    The contribution of a single value of xx to E(X(X−1))E(X\left(X-1\right)).

  15. State the value of E(X(X−1))E(X\left(X-1\right))

    E(X(X−1))=595E(X\left(X-1\right))=\frac{59}{5}

    This is the exact value of E(X(X−1))E(X\left(X-1\right)).

Answer
E(X(X−1))=595E(X\left(X-1\right))=\frac{59}{5}

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