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About Probability Distributions — Class 10 IB

Explore discrete probability distributions, expected value, and introduction to normal distribution. This topic is part of the IB Class 10 mathematics syllabus (chapter: Unit 8). On this page you can practice 58 questions across three difficulty levels — 20 easy, 18 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Probability Distributions — solved examples for Class 10 IB

Example 1easy

Which of the following statements correctly distinguishes a discrete random variable from a continuous random variable?
  1. A)A discrete random variable can take any value within a given range, while a continuous random variable can only take specific, isolated values.
  2. B)A discrete random variable has a finite or countably infinite number of possible values, while a continuous random variable can take any value within an interval.
  3. C)A discrete random variable is always represented by integers, whereas a continuous random variable is always represented by decimals.
  4. D)The probability distribution of a discrete random variable is always bell-shaped, while that of a continuous random variable is always uniform.

Step-by-step solution

  1. Step 1: Understand the definition of a discrete random variable. A discrete random variable can take on a finite or countably infinite number of distinct values, often integers, like the number of heads in coin flips or the number of cars passing a point.
  2. Step 2: Understand the definition of a continuous random variable. A continuous random variable can take any value within a given interval, often measured, like height, weight, or time.
  3. Step 3: Evaluate the given options based on these definitions. Option B accurately captures this fundamental distinction, as discrete variables have distinct, separate values, while continuous variables have an uncountably infinite number of values within any range.

Answer: A discrete random variable has a finite or countably infinite number of possible values, while a continuous random variable can take any value within an interval.

Example 2medium

Which of the following tables represents a valid probability distribution for a discrete random variable X?
  1. A)A) P(X=x): { (1, 0.2), (2, 0.3), (3, 0.6) }
  2. B)B) P(X=x): { (1, 0.1), (2, 0.4), (3, 0.5) }
  3. C)C) P(X=x): { (1, -0.1), (2, 0.6), (3, 0.5) }
  4. D)D) P(X=x): { (1, 0.3), (2, 0.4), (3, 0.2), (4, 0.2) }

Step-by-step solution

  1. For a probability distribution to be valid, two conditions must be met: 1) Each individual probability P(X=x) must be greater than or equal to 0. 2) The sum of all probabilities ΣP(X=x) must be equal to 1.
  2. A) ΣP(X=x) = 0.2 + 0.3 + 0.6 = 1.1. This is not equal to 1, so it's invalid.
  3. B) ΣP(X=x) = 0.1 + 0.4 + 0.5 = 1.0. All probabilities are non-negative. This is a valid distribution.
  4. C) P(X=1) = -0.1. Probabilities cannot be negative, so it's invalid.
  5. D) ΣP(X=x) = 0.3 + 0.4 + 0.2 + 0.2 = 1.1. This is not equal to 1, so it's invalid.

Answer: B) P(X=x): { (1, 0.1), (2, 0.4), (3, 0.5) }

Example 3hard

Consider a function P(x) defined for a discrete random variable X. Which of the following conditions, if violated, would *always* mean P(x) cannot be a valid probability mass function (PMF) for X?
  1. A)A. The sum of all P(x) values over the entire domain of X is less than 1.
  2. B)B. P(x) is negative for at least one value of x in the domain.
  3. C)C. The domain of X includes only positive integers.
  4. D)D. The expected value E(X) is not an integer.

Step-by-step solution

  1. A valid probability mass function (PMF) must satisfy two main conditions: (1) P(x) ≥ 0 for all x in the domain, and (2) ΣP(x) = 1 over the entire domain of X.
  2. Option A (sum less than 1) violates the second condition (ΣP(x) = 1), so it would not be a valid PMF. However, the question asks which violation would *always* mean P(x) cannot be a valid PMF, and if the sum is less than 1, it's not valid, but it's not the only way it can be invalid.
  3. Option B (P(x) is negative) directly violates the first and most fundamental condition that probabilities must be non-negative. A negative probability is impossible by definition.
  4. Option C (domain includes only positive integers) is not a general requirement for a PMF; the domain of a discrete random variable can include zero or negative integers. Option D (E(X) is not an integer) is also not a requirement; expected values can be non-integers.
  5. Therefore, P(x) being negative for any x is an absolute violation that makes it impossible to be a valid PMF.

Answer: B. P(x) is negative for at least one value of x in the domain.

Practice questions on Probability Distributions

  1. Q1.easy

    Ravi constructed a probability distribution for the number of successful free throws (X) out of two attempts for a basketball player. He listed the probabilities as P(X=0) = 0.2, P(X=1) = 0.5, and P(X=2) = 0.4. What error did Ravi make?
    1. A)The probabilities for a discrete random variable must always be equal.
    2. B)The sum of all probabilities in a probability distribution must equal 1.
    3. C)The probability for each outcome must be greater than 1.
    4. D)The number of outcomes (X values) must always be odd.
    Show answer

    Answer: The sum of all probabilities in a probability distribution must equal 1.

    Hint: Recall the fundamental properties that all valid probability distributions must satisfy. One key property involves the total probability.

  2. Q2.easy

    A fair six-sided die is rolled once. Let X be the random variable representing the number shown on the die. Which of the following correctly represents the probability distribution for X?
    1. A)A) P(X=x) = 1/6 for x ∈ {1, 2, 3, 4, 5, 6}
    2. B)B) P(X=x) = x/6 for x ∈ {1, 2, 3, 4, 5, 6}
    3. C)C) P(X=x) = 1/2 for x ∈ {1, 2, 3, 4, 5, 6}
    4. D)D) P(X=x) = 0.5 for x ∈ {1, 2, 3, 4, 5, 6}
    Show answer

    Answer: A) P(X=x) = 1/6 for x ∈ {1, 2, 3, 4, 5, 6}

    Hint: For a fair die, each outcome has an equal chance of occurring. Consider the total number of possible outcomes and the probability of each individual outcome.

  3. Q3.easy

    Consider a discrete random variable Y with the following probability distribution:

    | Y | 1 | 2 | 3 |
    |-------|-----|-----|-----|
    | P(Y=y) | 0.3 | 0.4 | 0.3 |

    Calculate the expected value, E(Y).
    1. A)2.0
    2. B)2.1
    3. C)2.2
    4. D)2.5
    Show answer

    Answer: 2.0

    Hint: Remember the formula for expected value: E(X) = Σx P(x). Multiply each value of Y by its corresponding probability and then sum these products.

  4. Q4.medium

    A discrete random variable Y has the following probability distribution: P(Y=y): { (10, 0.25), (20, p), (30, 0.15), (40, 0.3) }. Find the value of p.
    1. A)A) 0.1
    2. B)B) 0.2
    3. C)C) 0.3
    4. D)D) 0.4
    Show answer

    Answer: C) 0.3

    Hint: Remember that the sum of all probabilities for a discrete random variable must always be equal to 1.

  5. Q5.medium

    A fair six-sided die is rolled twice. Let X be the random variable representing the number of times a '6' appears. What is the expected value of X?
    1. A)A) 1/3
    2. B)B) 1/6
    3. C)C) 1/2
    4. D)D) 2/3
    Show answer

    Answer: A) 1/3

    Hint: First, determine the possible values for X and their corresponding probabilities. Then, use the formula for expected value: E(X) = Σ [x × P(X=x)].

  6. Q6.medium

    A carnival game involves spinning a wheel with 4 equally likely sectors: 'Win ₹100', 'Win ₹50', 'Lose ₹20', 'Lose ₹40'. If it costs ₹30 to play the game, what is the expected net gain (or loss) for a player?
    1. A)A) ₹5 loss
    2. B)B) ₹10 gain
    3. C)C) ₹10 loss
    4. D)D) ₹5 gain
    Show answer

    Answer: A) ₹5 loss

    Hint: First, calculate the actual monetary outcome for each sector after accounting for the game cost. Then, find the expected value of these net outcomes.

  7. Q7.hard

    A discrete random variable X has the following probability mass function: P(X=x) = k(x² + 1) for x = 0, 1, 2, 3, and P(X=x) = 0 otherwise. If E(X) = 2.05, what is the value of P(X > 1 | X < 3)?
    1. A)A. 13/20
    2. B)B. 10/17
    3. C)C. 13/17
    4. D)D. 10/20
    Show answer

    Answer: B. 10/17

    Hint: First, use the property that the sum of all probabilities must equal 1 to find the value of k. Then, compute the expected value using the formula E(X) = ΣxP(x) to verify k, or if E(X) is given, use it to find k. Finally, apply the conditional probability formula P(A|B) = P(A ∩ B) / P(B).

  8. Q8.hard

    A game involves rolling a fair six-sided die twice. Let X be the random variable representing the number of times a 'prime number' (2, 3, or 5) appears in the two rolls. What is the expected value of X, E(X)?
    1. A)A. 1
    2. B)B. 2/3
    3. C)C. 4/3
    4. D)D. 3/2
    Show answer

    Answer: A. 1

    Hint: First, determine the probability of rolling a prime number in a single roll. Then, identify the possible values for X and calculate their probabilities using binomial probability concepts for two independent trials. Finally, apply the formula for expected value.

  9. Q9.hard

    A company produces light bulbs. The lifespan of a bulb, in hours, is a random variable X. Due to a manufacturing defect, 10% of the bulbs have a lifespan of 500 hours, 40% have a lifespan of 1000 hours, and 50% have a lifespan of 2000 hours. If a customer buys two such bulbs, and their lifespans are independent, what is the expected sum of the lifespans of the two bulbs?
    1. A)A. 1550 hours
    2. B)B. 1550000 hours
    3. C)C. 3100 hours
    4. D)D. 2000 hours
    Show answer

    Answer: C. 3100 hours

    Hint: First, calculate the expected lifespan of a single bulb. Remember that for independent random variables X₁ and X₂, E(X₁ + X₂) = E(X₁) + E(X₂).

These are 9 of the 58 questions available for Probability Distributions. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.