Hard A-Level Discrete random variables Questions

Challenging, exam-style A-Level Discrete random variables questions with worked solutions. Stretch yourself on the hardest probability distribution, probabilities sum to 1, cumulative probability, valid distribution problems.

probability distributionprobabilities sum to 1cumulative probabilityvalid distributionprobability function
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A discrete random variable XX has a cubic weighting. The discrete random variable XX has probability function P(X=x)=kx3P(X=x)=kx^3 for x=1,2,3,4x=1,2,3,4, and P(X=x)=0P(X=x)=0 otherwise, where kk is a constant. Find the value of kk, and hence find P(X≤2)P(X \le 2), P(X≥3)P(X \ge 3) and P(2≤X≤4)P(2 \le X \le 4).
Show worked solution

Worked solution

  1. Use that all the probabilities sum to 1

    ∑P(X=x)=1\sum P(X=x) = 1

    The probability function must give probabilities that total 1.

  2. Substitute the probability function over its range

    k(1+8+27+64)=1k(1 + 8 + 27 + 64) = 1

    Add the value of the function at each value of x.

  3. Add the values inside the bracket

    100k=1100k = 1

    Combine the terms to get a single equation in k.

  4. Solve the equation for k

    k=1100k = \frac{1}{100}

    Divide to obtain the unknown constant.

  5. Tabulate the probability P(X=1)P(X=1)

    P(X=1)=1100×1=1100P(X=1) = \frac{1}{100} \times 1 = \frac{1}{100}

    Substitute this value of x into the probability function.

  6. Tabulate the probability P(X=2)P(X=2)

    P(X=2)=1100×8=225P(X=2) = \frac{1}{100} \times 8 = \frac{2}{25}

    Substitute this value of x into the probability function.

  7. Tabulate the probability P(X=3)P(X=3)

    P(X=3)=1100×27=27100P(X=3) = \frac{1}{100} \times 27 = \frac{27}{100}

    Substitute this value of x into the probability function.

  8. Tabulate the probability P(X=4)P(X=4)

    P(X=4)=1100×64=1625P(X=4) = \frac{1}{100} \times 64 = \frac{16}{25}

    Substitute this value of x into the probability function.

  9. Write P(X≤2)P(X \le 2) as a sum of outcomes

    P(X≤2)=P(X=1)+P(X=2)P(X \le 2) = P(X=1) + P(X=2)

    Include every value of X that satisfies the condition.

  10. Substitute the probabilities and add

    =1100+225=9100= \frac{1}{100} + \frac{2}{25} = \frac{9}{100}

    Add the probabilities of the relevant outcomes.

  11. Write P(X≥3)P(X \ge 3) as a sum of outcomes

    P(X≥3)=P(X=3)+P(X=4)P(X \ge 3) = P(X=3) + P(X=4)

    Include every value of X that satisfies the condition.

  12. Substitute the probabilities and add

    =27100+1625=91100= \frac{27}{100} + \frac{16}{25} = \frac{91}{100}

    Add the probabilities of the relevant outcomes.

  13. Write P(2≤X≤4)P(2 \le X \le 4) as a sum of outcomes

    P(2≤X≤4)=P(X=2)+P(X=3)+P(X=4)P(2 \le X \le 4) = P(X=2) + P(X=3) + P(X=4)

    Include every value of X that satisfies the condition.

  14. Substitute the probabilities and add

    =225+27100+1625=99100= \frac{2}{25} + \frac{27}{100} + \frac{16}{25} = \frac{99}{100}

    Add the probabilities of the relevant outcomes.

  15. Confirm the distribution is complete

    F(4)=P(X≤4)=1F(4) = P(X \le 4) = 1

    The cumulative probability reaches 1 at the largest value, as required.

Answer
k=1100; P(X≤2)=9100; P(X≥3)=91100; P(2≤X≤4)=99100k = \frac{1}{100};\ P(X \le 2) = \frac{9}{100};\ P(X \ge 3) = \frac{91}{100};\ P(2 \le X \le 4) = \frac{99}{100}
Question 2
8 markschallenging
A discrete random variable XX has a quadratic-plus rule. The discrete random variable XX has probability function P(X=x)=k(x2+1)P(X=x)=k(x^2+1) for x=1,2,3,4x=1,2,3,4, and P(X=x)=0P(X=x)=0 otherwise, where kk is a constant. Find the value of kk, and hence find P(X≤2)P(X \le 2), P(X>3)P(X > 3) and P(1<X≤2)P(1 < X \le 2).
Show worked solution

Worked solution

  1. Use that all the probabilities sum to 1

    ∑P(X=x)=1\sum P(X=x) = 1

    The probability function must give probabilities that total 1.

  2. Substitute the probability function over its range

    k(2+5+10+17)=1k(2 + 5 + 10 + 17) = 1

    Add the value of the function at each value of x.

  3. Add the values inside the bracket

    34k=134k = 1

    Combine the terms to get a single equation in k.

  4. Solve the equation for k

    k=134k = \frac{1}{34}

    Divide to obtain the unknown constant.

  5. Tabulate the probability P(X=1)P(X=1)

    P(X=1)=134×2=117P(X=1) = \frac{1}{34} \times 2 = \frac{1}{17}

    Substitute this value of x into the probability function.

  6. Tabulate the probability P(X=2)P(X=2)

    P(X=2)=134×5=534P(X=2) = \frac{1}{34} \times 5 = \frac{5}{34}

    Substitute this value of x into the probability function.

  7. Tabulate the probability P(X=3)P(X=3)

    P(X=3)=134×10=517P(X=3) = \frac{1}{34} \times 10 = \frac{5}{17}

    Substitute this value of x into the probability function.

  8. Tabulate the probability P(X=4)P(X=4)

    P(X=4)=134×17=12P(X=4) = \frac{1}{34} \times 17 = \frac{1}{2}

    Substitute this value of x into the probability function.

  9. Write P(X≤2)P(X \le 2) as a sum of outcomes

    P(X≤2)=P(X=1)+P(X=2)P(X \le 2) = P(X=1) + P(X=2)

    Include every value of X that satisfies the condition.

  10. Substitute the probabilities and add

    =117+534=734= \frac{1}{17} + \frac{5}{34} = \frac{7}{34}

    Add the probabilities of the relevant outcomes.

  11. Locate the outcome for P(X>3)P(X > 3)

    P(X>3)=P(X=4)P(X > 3) = P(X=4)

    Identify which listed value of X meets the stated condition.

  12. State the required probability

    P(X>3)=12P(X > 3) = \frac{1}{2}

    Read the matching probability straight from the distribution.

  13. Locate the outcome for P(1<X≤2)P(1 < X \le 2)

    P(1<X≤2)=P(X=2)P(1 < X \le 2) = P(X=2)

    Identify which listed value of X meets the stated condition.

  14. State the required probability

    P(1<X≤2)=534P(1 < X \le 2) = \frac{5}{34}

    Read the matching probability straight from the distribution.

  15. Confirm the distribution is complete

    F(4)=P(X≤4)=1F(4) = P(X \le 4) = 1

    The cumulative probability reaches 1 at the largest value, as required.

Answer
k=134; P(X≤2)=734; P(X>3)=12; P(1<X≤2)=534k = \frac{1}{34};\ P(X \le 2) = \frac{7}{34};\ P(X > 3) = \frac{1}{2};\ P(1 < X \le 2) = \frac{5}{34}
Question 3
8 markschallenging
A random variable XX follows a linear scoring rule. The discrete random variable XX has probability function P(X=x)=k(3x−2)P(X=x)=k(3x-2) for x=1,2,3,4x=1,2,3,4, and P(X=x)=0P(X=x)=0 otherwise, where kk is a constant. Find the value of kk, and hence find P(X≤3)P(X \le 3), P(X>2)P(X > 2) and P(2≤X≤3)P(2 \le X \le 3).
Show worked solution

Worked solution

  1. Use that all the probabilities sum to 1

    ∑P(X=x)=1\sum P(X=x) = 1

    The probability function must give probabilities that total 1.

  2. Substitute the probability function over its range

    k(1+4+7+10)=1k(1 + 4 + 7 + 10) = 1

    Add the value of the function at each value of x.

  3. Add the values inside the bracket

    22k=122k = 1

    Combine the terms to get a single equation in k.

  4. Solve the equation for k

    k=122k = \frac{1}{22}

    Divide to obtain the unknown constant.

  5. Tabulate the probability P(X=1)P(X=1)

    P(X=1)=122×1=122P(X=1) = \frac{1}{22} \times 1 = \frac{1}{22}

    Substitute this value of x into the probability function.

  6. Tabulate the probability P(X=2)P(X=2)

    P(X=2)=122×4=211P(X=2) = \frac{1}{22} \times 4 = \frac{2}{11}

    Substitute this value of x into the probability function.

  7. Tabulate the probability P(X=3)P(X=3)

    P(X=3)=122×7=722P(X=3) = \frac{1}{22} \times 7 = \frac{7}{22}

    Substitute this value of x into the probability function.

  8. Tabulate the probability P(X=4)P(X=4)

    P(X=4)=122×10=511P(X=4) = \frac{1}{22} \times 10 = \frac{5}{11}

    Substitute this value of x into the probability function.

  9. Write P(X≤3)P(X \le 3) as a sum of outcomes

    P(X≤3)=P(X=1)+P(X=2)+P(X=3)P(X \le 3) = P(X=1) + P(X=2) + P(X=3)

    Include every value of X that satisfies the condition.

  10. Substitute the probabilities and add

    =122+211+722=611= \frac{1}{22} + \frac{2}{11} + \frac{7}{22} = \frac{6}{11}

    Add the probabilities of the relevant outcomes.

  11. Write P(X>2)P(X > 2) as a sum of outcomes

    P(X>2)=P(X=3)+P(X=4)P(X > 2) = P(X=3) + P(X=4)

    Include every value of X that satisfies the condition.

  12. Substitute the probabilities and add

    =722+511=1722= \frac{7}{22} + \frac{5}{11} = \frac{17}{22}

    Add the probabilities of the relevant outcomes.

  13. Write P(2≤X≤3)P(2 \le X \le 3) as a sum of outcomes

    P(2≤X≤3)=P(X=2)+P(X=3)P(2 \le X \le 3) = P(X=2) + P(X=3)

    Include every value of X that satisfies the condition.

  14. Substitute the probabilities and add

    =211+722=12= \frac{2}{11} + \frac{7}{22} = \frac{1}{2}

    Add the probabilities of the relevant outcomes.

  15. Confirm the distribution is complete

    F(4)=P(X≤4)=1F(4) = P(X \le 4) = 1

    The cumulative probability reaches 1 at the largest value, as required.

Answer
k=122; P(X≤3)=611; P(X>2)=1722; P(2≤X≤3)=12k = \frac{1}{22};\ P(X \le 3) = \frac{6}{11};\ P(X > 2) = \frac{17}{22};\ P(2 \le X \le 3) = \frac{1}{2}
Question 4
8 markschallenging
A random variable XX uses a linear weighting. The discrete random variable XX has probability function P(X=x)=k(x+3)P(X=x)=k(x+3) for x=1,2,3,4x=1,2,3,4, and P(X=x)=0P(X=x)=0 otherwise, where kk is a constant. Find the value of kk, and hence find P(X≤2)P(X \le 2), P(X≥3)P(X \ge 3) and P(1<X≤3)P(1 < X \le 3).
Show worked solution

Worked solution

  1. Use that all the probabilities sum to 1

    ∑P(X=x)=1\sum P(X=x) = 1

    The probability function must give probabilities that total 1.

  2. Substitute the probability function over its range

    k(4+5+6+7)=1k(4 + 5 + 6 + 7) = 1

    Add the value of the function at each value of x.

  3. Add the values inside the bracket

    22k=122k = 1

    Combine the terms to get a single equation in k.

  4. Solve the equation for k

    k=122k = \frac{1}{22}

    Divide to obtain the unknown constant.

  5. Tabulate the probability P(X=1)P(X=1)

    P(X=1)=122×4=211P(X=1) = \frac{1}{22} \times 4 = \frac{2}{11}

    Substitute this value of x into the probability function.

  6. Tabulate the probability P(X=2)P(X=2)

    P(X=2)=122×5=522P(X=2) = \frac{1}{22} \times 5 = \frac{5}{22}

    Substitute this value of x into the probability function.

  7. Tabulate the probability P(X=3)P(X=3)

    P(X=3)=122×6=311P(X=3) = \frac{1}{22} \times 6 = \frac{3}{11}

    Substitute this value of x into the probability function.

  8. Tabulate the probability P(X=4)P(X=4)

    P(X=4)=122×7=722P(X=4) = \frac{1}{22} \times 7 = \frac{7}{22}

    Substitute this value of x into the probability function.

  9. Write P(X≤2)P(X \le 2) as a sum of outcomes

    P(X≤2)=P(X=1)+P(X=2)P(X \le 2) = P(X=1) + P(X=2)

    Include every value of X that satisfies the condition.

  10. Substitute the probabilities and add

    =211+522=922= \frac{2}{11} + \frac{5}{22} = \frac{9}{22}

    Add the probabilities of the relevant outcomes.

  11. Write P(X≥3)P(X \ge 3) as a sum of outcomes

    P(X≥3)=P(X=3)+P(X=4)P(X \ge 3) = P(X=3) + P(X=4)

    Include every value of X that satisfies the condition.

  12. Substitute the probabilities and add

    =311+722=1322= \frac{3}{11} + \frac{7}{22} = \frac{13}{22}

    Add the probabilities of the relevant outcomes.

  13. Write P(1<X≤3)P(1 < X \le 3) as a sum of outcomes

    P(1<X≤3)=P(X=2)+P(X=3)P(1 < X \le 3) = P(X=2) + P(X=3)

    Include every value of X that satisfies the condition.

  14. Substitute the probabilities and add

    =522+311=12= \frac{5}{22} + \frac{3}{11} = \frac{1}{2}

    Add the probabilities of the relevant outcomes.

  15. Confirm the distribution is complete

    F(4)=P(X≤4)=1F(4) = P(X \le 4) = 1

    The cumulative probability reaches 1 at the largest value, as required.

Answer
k=122; P(X≤2)=922; P(X≥3)=1322; P(1<X≤3)=12k = \frac{1}{22};\ P(X \le 2) = \frac{9}{22};\ P(X \ge 3) = \frac{13}{22};\ P(1 < X \le 3) = \frac{1}{2}
Question 5
8 markschallenging
A discrete random variable XX is defined on higher values. The discrete random variable XX has probability function P(X=x)=kxP(X=x)=kx for x=3,4,5,6x=3,4,5,6, and P(X=x)=0P(X=x)=0 otherwise, where kk is a constant. Find the value of kk, and hence find P(X≤4)P(X \le 4), P(X>5)P(X > 5) and P(4≤X≤6)P(4 \le X \le 6).
Show worked solution

Worked solution

  1. Use that all the probabilities sum to 1

    ∑P(X=x)=1\sum P(X=x) = 1

    The probability function must give probabilities that total 1.

  2. Substitute the probability function over its range

    k(3+4+5+6)=1k(3 + 4 + 5 + 6) = 1

    Add the value of the function at each value of x.

  3. Add the values inside the bracket

    18k=118k = 1

    Combine the terms to get a single equation in k.

  4. Solve the equation for k

    k=118k = \frac{1}{18}

    Divide to obtain the unknown constant.

  5. Tabulate the probability P(X=3)P(X=3)

    P(X=3)=118×3=16P(X=3) = \frac{1}{18} \times 3 = \frac{1}{6}

    Substitute this value of x into the probability function.

  6. Tabulate the probability P(X=4)P(X=4)

    P(X=4)=118×4=29P(X=4) = \frac{1}{18} \times 4 = \frac{2}{9}

    Substitute this value of x into the probability function.

  7. Tabulate the probability P(X=5)P(X=5)

    P(X=5)=118×5=518P(X=5) = \frac{1}{18} \times 5 = \frac{5}{18}

    Substitute this value of x into the probability function.

  8. Tabulate the probability P(X=6)P(X=6)

    P(X=6)=118×6=13P(X=6) = \frac{1}{18} \times 6 = \frac{1}{3}

    Substitute this value of x into the probability function.

  9. Write P(X≤4)P(X \le 4) as a sum of outcomes

    P(X≤4)=P(X=3)+P(X=4)P(X \le 4) = P(X=3) + P(X=4)

    Include every value of X that satisfies the condition.

  10. Substitute the probabilities and add

    =16+29=718= \frac{1}{6} + \frac{2}{9} = \frac{7}{18}

    Add the probabilities of the relevant outcomes.

  11. Locate the outcome for P(X>5)P(X > 5)

    P(X>5)=P(X=6)P(X > 5) = P(X=6)

    Identify which listed value of X meets the stated condition.

  12. State the required probability

    P(X>5)=13P(X > 5) = \frac{1}{3}

    Read the matching probability straight from the distribution.

  13. Write P(4≤X≤6)P(4 \le X \le 6) as a sum of outcomes

    P(4≤X≤6)=P(X=4)+P(X=5)+P(X=6)P(4 \le X \le 6) = P(X=4) + P(X=5) + P(X=6)

    Include every value of X that satisfies the condition.

  14. Substitute the probabilities and add

    =29+518+13=56= \frac{2}{9} + \frac{5}{18} + \frac{1}{3} = \frac{5}{6}

    Add the probabilities of the relevant outcomes.

  15. Confirm the distribution is complete

    F(6)=P(X≤6)=1F(6) = P(X \le 6) = 1

    The cumulative probability reaches 1 at the largest value, as required.

Answer
k=118; P(X≤4)=718; P(X>5)=13; P(4≤X≤6)=56k = \frac{1}{18};\ P(X \le 4) = \frac{7}{18};\ P(X > 5) = \frac{1}{3};\ P(4 \le X \le 6) = \frac{5}{6}

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