DP Math AA · HL / SL · Statistics & Probability

SL 4.7—Discrete random variables

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  1. Question 1

    The discrete random variable X has the following probability distribution:

    x123
    P(X=x)3kk0.3

    Find the value of k.

    No clue? Show me the answer
    Correct answerCorrect!Incorrect
    Ak=0.175

    Step-by-step walkthrough

    Choose a solution method

    Method #1Approach 1

    Step 1: State the condition for a valid probability distribution

    All probabilities must sum to 1: ∑P(X=x)=1, so 3k+k+0.3=1.

    Step 2: Simplify and solve for $k$

    Combining like terms: 4k+0.3=1, giving 4k=0.7, so k=0.175.

    Step 3: Verify the solution

    Check: 3(0.175)+0.175+0.3=0.525+0.175+0.3=1 ✓. The answer is k=0.175.

    Method #2Approach 2

    Step 1: Identify what is needed

    We need 3k+k+0.3=1, i.e. 4k=0.7, so k=0.175.

    Step 2: Eliminate $k = 0.7$

    If k=0.7, then 3(0.7)+0.7+0.3=2.1+0.7+0.3=3.1=1. This is invalid.

    Step 3: Eliminate $k = 0.25$

    If k=0.25, then 3(0.25)+0.25+0.3=0.75+0.25+0.3=1.3=1. This is invalid.

    Step 4: Eliminate $k = 0.1$

    If k=0.1, then 3(0.1)+0.1+0.3=0.3+0.1+0.3=0.7=1. This is invalid.

    Step 5: Select the correct answer

    Only k=0.175 satisfies 4k=0.7, giving probabilities that sum to 1.

  2. Question 2

    A discrete random variable W has the probability distribution shown below:

    w0123
    P(W=w)0.150.40m0.05

    What is the value of m?

    No clue? Show me the answer
    Correct answerCorrect!Incorrect
    Am=0.40

    Step-by-step walkthrough

    Choose a solution method

    Method #1Approach 1

    Step 1: Use the sum-to-one rule

    For any valid probability distribution, all probabilities must sum to 1: 0.15+0.40+m+0.05=1.

    Step 2: Solve for $m$

    Adding the known values: 0.60+m=1, so m=0.40.

    Step 3: Confirm validity

    Since 0≤0.40≤1, this is a valid probability. The answer is m=0.40.

    Method #2Approach 2

    Step 1: Identify the constraint

    The known probabilities sum to 0.15+0.40+0.05=0.60, so m=1−0.60=0.40.

    Step 2: Eliminate $m = 0.35$

    If m=0.35, total =0.60+0.35=0.95=1. Not valid.

    Step 3: Eliminate $m = 0.45$

    If m=0.45, total =0.60+0.45=1.05=1. Exceeds 1, so invalid.

    Step 4: Eliminate $m = 0.30$

    If m=0.30, total =0.60+0.30=0.90=1. Not valid.

    Step 5: Select the correct answer

    Only m=0.40 makes the probabilities sum to exactly 1.

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← Previous topicSL 4.6—Combined, mutually exclusive, conditional, independence, prob diagramsNext topic →SL 4.8—Binomial distribution
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