A-Level Discrete random variables Practice Questions

Free A-Level Discrete random variables practice questions with full step-by-step worked solutions. Covers discrete random variable, probability distribution, probabilities sum to 1, cumulative probability. Practise exam-style problems and check your method.

discrete random variableprobability distributionprobabilities sum to 1cumulative probabilitydiscrete uniform distributionprobability
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A discrete random variable XX has the probability distribution given by P(X=0)=0.1P(X=0)=0.1, P(X=1)=0.3P(X=1)=0.3, P(X=2)=kP(X=2)=k, P(X=3)=0.2P(X=3)=0.2. Find the value of kk.
Show worked solution

Worked solution

  1. Use that all the probabilities sum to 1

    ∑P(X=x)=1\sum P(X=x) = 1

    For a discrete random variable the probabilities of every outcome add to 1.

  2. Substitute the entries of the distribution

    0.1+0.3+k+0.2=10.1 + 0.3 + k + 0.2 = 1

    Write the total of all the listed probabilities.

  3. Solve the equation for k

    k=0.4k = 0.4

    Rearrange to find the unknown constant.

Answer
k=0.4k = 0.4
Question 2
2 markseasy
Which of the following can NOT be a probability distribution?
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Worked solution

  1. Recall the conditions a distribution must satisfy

    0≤P(X=x)≤1, ∑P(X=x)=10 \le P(X=x) \le 1,\ \sum P(X=x) = 1

    Probabilities cannot be negative and must total 1.

  2. Scan the options for a rule that is broken

    P(X=2)=−0.2<0P(X=2) = -0.2 < 0

    A negative probability is impossible.

  3. Select the impossible distribution

    P(X=x)<0 is not allowedP(X=x) < 0 \text{ is not allowed}

    The option containing a negative probability cannot be a distribution.

Answer
P(X=1)=0.6,P(X=2)=−0.2,P(X=3)=0.6P(X=1)=0.6, P(X=2)=-0.2, P(X=3)=0.6
Question 3
3 marksintermediate
For a discrete random variable XX, which outcomes are counted by P(2<X≤5)P(2 < X \le 5)?
Show worked solution

Worked solution

  1. Read the lower bound

    2<X2 < X

    X must be strictly greater than 2, so 2 is excluded.

  2. Read the upper bound

    X≤5X \le 5

    X is at most 5, so 5 is included.

  3. Combine the two conditions

    2<X≤52 < X \le 5

    X lies strictly above 2 and up to 5.

  4. List integer outcomes in range

    X∈{3,4,5}X \in \{3, 4, 5\}

    These are the whole-number values that qualify.

  5. Express as a sum of probabilities

    P(X=3)+P(X=4)+P(X=5)P(X=3) + P(X=4) + P(X=5)

    Add the probabilities of the included values.

  6. Select the correct set

    {3,4,5}\{3, 4, 5\}

    The values 3, 4 and 5 are counted.

Answer
X=3,4,5X = 3, 4, 5
Question 4
5 markshard
A discrete random variable XX has the probability distribution given by P(X=1)=0.2P(X=1)=0.2, P(X=2)=kP(X=2)=k, P(X=3)=0.3P(X=3)=0.3, P(X=4)=0.2P(X=4)=0.2. Which value below is correct for this distribution?
Show worked solution

Worked solution

  1. Use that all the probabilities sum to 1

    ∑P(X=x)=1\sum P(X=x) = 1

    For a discrete random variable the probabilities of every outcome add to 1.

  2. Substitute the entries of the distribution

    0.2+k+0.3+0.2=10.2 + k + 0.3 + 0.2 = 1

    Write the total of all the listed probabilities.

  3. Collect the terms of the equation

    k+0.7=1k + 0.7 = 1

    Group the known probabilities and the terms in k.

  4. Solve the equation for k

    k=0.3k = 0.3

    Rearrange to find the unknown constant.

  5. 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.

  6. Substitute the probabilities and add

    =0.2+0.3=0.5= 0.2 + 0.3 = 0.5

    Add the probabilities of the relevant outcomes.

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

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

    Include every value of X that satisfies the condition.

  8. Substitute the probabilities and add

    =0.3+0.2=0.5= 0.3 + 0.2 = 0.5

    Add the probabilities of the relevant outcomes.

  9. 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.

  10. Substitute the probabilities and add

    =0.3+0.2=0.5= 0.3 + 0.2 = 0.5

    Add the probabilities of the relevant outcomes.

Answer
P(2<X≤4)=0.5P(2 < X \le 4) = 0.5
Question 5
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}

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