Practice Maths

Discrete Random Variables and Probability Functions — Full Worked Solutions

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Full Worked Solutions

  1. Question: Show that the following is a valid probability function for X: x = 0, 1, 2, 3 with P(X=x) = 0.1, 0.4, 0.3, 0.2.

    Condition 1 — Non-negativity

    All probabilities: 0.1, 0.4, 0.3, 0.2 are all ≥ 0. ✓

    Condition 2 — Probabilities Sum to 1

    0.1 + 0.4 + 0.3 + 0.2 = 1.0 ✓

    Both conditions hold, so this is a valid probability function.

  2. Question: X has probability function P(X = x) = k(x + 1) for x = 0, 1, 2, 3. Find k.

    Apply ∑ P(X = x) = 1

    k(0+1) + k(1+1) + k(2+1) + k(3+1) = 1
    k(1 + 2 + 3 + 4) = 1
    10k = 1
    k = 1/10 = 0.1

    Verification

    P(X=0) = 0.1, P(X=1) = 0.2, P(X=2) = 0.3, P(X=3) = 0.4. Sum = 1.0 ✓

  3. Question: For the distribution with x = 1, 2, 3, 4 and P(X=x) = 0.25, 0.30, p, 0.15, find the missing probability p and then find P(X ≥ 3).

    Find p

    Using ∑ P(X = x) = 1:
    0.25 + 0.30 + p + 0.15 = 1
    0.70 + p = 1
    p = 0.30

    Find P(X ≥ 3)

    P(X ≥ 3) = P(X=3) + P(X=4) = 0.30 + 0.15 = 0.45

  4. Question: Using the distribution from Q3 (p = 0.30), find P(X > 2) and P(1 ≤ X ≤ 3).

    P(X > 2)

    Method 1 (complement): P(X > 2) = 1 − P(X ≤ 2) = 1 − (0.25 + 0.30) = 1 − 0.55 = 0.45
    Method 2 (direct): P(X=3) + P(X=4) = 0.30 + 0.15 = 0.45 ✓

    P(1 ≤ X ≤ 3)

    P(X=1) + P(X=2) + P(X=3) = 0.25 + 0.30 + 0.30 = 0.85

  5. Question: Which of the following is a valid probability function for X with values {1, 2, 3}? (A) P(X=x)=x/4; (B) P(X=x)=(4−x)/9; (C) P(X=x)=(x−1)/3. Justify your answer.

    Check (A): P(X=x) = x/4

    P(X=1) = 1/4, P(X=2) = 2/4, P(X=3) = 3/4
    Sum = 1/4 + 2/4 + 3/4 = 6/4 = 1.5 ≠ 1. Not valid.

    Check (B): P(X=x) = (4−x)/9

    P(X=1) = 3/9, P(X=2) = 2/9, P(X=3) = 1/9
    Sum = 6/9 = 2/3 ≠ 1. Not valid.

    Check (C): P(X=x) = (x−1)/3

    P(X=1) = 0, P(X=2) = 1/3, P(X=3) = 2/3
    All values ≥ 0 ✓
    Sum = 0 + 1/3 + 2/3 = 1 ✓
    Option (C) is the valid probability function.

  6. Question: A probability function is given by P(X = x) = c/x for x = 1, 2, 3, 4, 5, 6. Find c. Then construct the probability distribution table and calculate P(3 ≤ X ≤ 5).

    Find c

    ∑ P(X = x) = c(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) = 1
    Convert to 60ths: c × (60 + 30 + 20 + 15 + 12 + 10)/60 = 1
    c × 147/60 = 1
    c = 60/147 = 20/49

    Probability Distribution Table

    P(X=1) = 20/49 ≈ 0.408
    P(X=2) = 10/49 ≈ 0.204
    P(X=3) = 20/147 ≈ 0.136
    P(X=4) = 5/49 ≈ 0.102
    P(X=5) = 4/49 ≈ 0.082
    P(X=6) = 10/147 ≈ 0.068

    P(3 ≤ X ≤ 5)

    = 20/147 + 5/49 + 4/49
    = 20/147 + 15/147 + 12/147
    = 47/147 ≈ 0.320

  7. Question: A bag contains 3 red, 2 blue, and 1 green marble. One marble is drawn. Let X = 1 if red, X = 2 if blue, X = 3 if green. Construct the probability distribution table for X and find P(X ≠ 2).

    Building the Distribution

    Total marbles = 6.
    P(X=1) = 3/6 = 1/2 (3 red marbles)
    P(X=2) = 2/6 = 1/3 (2 blue marbles)
    P(X=3) = 1/6 (1 green marble)

    Verify: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1 ✓

    P(X ≠ 2)

    Using the complement rule:
    P(X ≠ 2) = 1 − P(X=2) = 1 − 1/3 = 2/3

  8. Question: X has distribution: x = −1, 0, 1, 2 with P(X=x) = 0.20, 0.35, 0.30, 0.15. Find (a) P(X < 1)   (b) P(−1 ≤ X ≤ 1)   (c) P(X ≥ 0).

    (a) P(X < 1)

    X < 1 means X = −1 or X = 0:
    P(X < 1) = P(X=−1) + P(X=0) = 0.20 + 0.35 = 0.55

    (b) P(−1 ≤ X ≤ 1)

    = P(X=−1) + P(X=0) + P(X=1) = 0.20 + 0.35 + 0.30 = 0.85

    (c) P(X ≥ 0)

    = P(X=0) + P(X=1) + P(X=2) = 0.35 + 0.30 + 0.15 = 0.80
    Or via complement: 1 − P(X=−1) = 1 − 0.20 = 0.80 ✓

  9. Question: X has probability function P(X = x) = k(2x − 1) for x = 1, 2, 3, 4. Find k, state the probability distribution, identify the mode, and find P(X is even).

    Find k

    P(X=1) = k(1), P(X=2) = k(3), P(X=3) = k(5), P(X=4) = k(7)
    Apply ∑ = 1: k(1 + 3 + 5 + 7) = 16k = 1
    k = 1/16

    Probability Distribution

    P(X=1) = 1/16, P(X=2) = 3/16, P(X=3) = 5/16, P(X=4) = 7/16

    Mode

    The mode is the value with the highest probability. P(X=4) = 7/16 is the largest.
    Mode = 4

    P(X is even)

    P(X is even) = P(X=2) + P(X=4) = 3/16 + 7/16 = 10/16 = 5/8 = 0.625

  10. Question: Two fair coins are tossed simultaneously. Let X be the number of heads. (a) Construct the probability distribution table for X. (b) Find P(X ≥ 1). (c) Explain why P(X=0) + P(X=1) + P(X=2) = 1 must hold.

    (a) Probability Distribution

    Sample space: {HH, HT, TH, TT} — 4 equally likely outcomes.
    P(X=0) = 1/4   (outcome: TT)
    P(X=1) = 2/4 = 1/2   (outcomes: HT, TH)
    P(X=2) = 1/4   (outcome: HH)

    Verify: 1/4 + 1/2 + 1/4 = 1 ✓

    (b) P(X ≥ 1)

    Using the complement:
    P(X ≥ 1) = 1 − P(X=0) = 1 − 1/4 = 3/4

    (c) Why the Probabilities Must Sum to 1

    The values {0, 1, 2} represent all possible numbers of heads when two coins are tossed. These outcomes are mutually exclusive (you cannot get both 0 and 1 heads simultaneously) and exhaustive (one of these outcomes must occur on every toss). By the total probability rule, the sum of probabilities of all possible outcomes equals 1, since one outcome is guaranteed to occur.