Curriculum
Course: Statistics — Third Secondary
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Curriculum

Statistics — Third Secondary

Unit 1: Correlation and Regression

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Unit 2: Advanced Statistical Measures

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Unit 3: Probability

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Unit 4: Random Variables

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Unit 5: The Normal Distribution

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Text lesson

Discrete Random Variables

Statistics — Third Secondary

Discrete Random Variables

Worked examples

Worked examples

Example 1→Two fair coins are tossed, and X is the number of heads. Write the probability distribution of X and check its conditions.
Comparison aspectMain ideaKeywords
The sample space has 4 equally likely outcomes
•
  • X = 0 in one outcome (TT)
  • X = 1 in two outcomes (HT
  • TH)
  • and X = 2 in one outcome (HH)
outcomeoutcomes
Distribution
  • f(0) = 1 ÷ 4
  • f(1) = 2 ÷ 4 = 1 ÷ 2
  • f(2) = 1 ÷ 4
•
  • Sum of the probabilities = 1 ÷ 4 + 1 ÷ 2 + 1 ÷ 4 = 1
  • so it is a valid distribution
probabilitiesvaliddistribution
Example 2
  • A random variable X takes the values 1
  • with probability f(x) = k × x. Find the value of k
  • then find P(X ≥ 3) and P(X is even)
randomvariabletakesvalues
Comparison aspectMain ideaKeywords
Sum of the probabilities
  • k(1 + 2 + 3 + 4) = 10k = 1
  • so k = 1 ÷ 10
Distribution
  • f(1) = 0.1
  • f(2) = 0.2
  • f(3) = 0.3
  • f(4) = 0.4
•P(X ≥ 3) = 0.3 + 0.4 = 0.7
•P(X is even) = P(X = 2) + P(X = 4) = 0.2 + 0.4 = 0.6even
Example 3
  • The distribution of X is: X = 0
  • 3 with probabilities 0.1
  • 0.3
  • 0.2 respectively. Find a
  • then find P(X ≤ 1) and P(1 ≤ X < 3)
distributionprobabilitiesrespectivelyFind
  • 0.1 + a + 0.3 + 0.2 = 1, so a = 0.4.
  • P(X ≤ 1) = f(0) + f(1) = 0.1 + 0.4 = 0.5.
  • P(1 ≤ X < 3) = f(1) + f(2) = 0.4 + 0.3 = 0.7, because the value 3 is not included in the event.
  • Example 4 A fair six-sided die is rolled twice, and X is the number of times the number 6 appears. Write the probability distribution of X, then find P(X ≥ 1).
  • The number of outcomes of the experiment is 6 × 6 = 36, all equally likely.
  • X = 0 no 6 appears in either roll. The number of outcomes is 5 × 5 = 25, so f(0) = 25 ÷ 36.
  • X = 1 a 6 appears in exactly one roll. The number of outcomes is 2 × 5 = 10, so f(1) = 10 ÷ 36.
  • X = 2 a 6 appears in both rolls. The number of outcomes is 1, so f(2) = 1 ÷ 36.
  • Sum of the probabilities = (25 + 10 + 1) ÷ 36 = 1.
  • P(X ≥ 1) = 10 ÷ 36 + 1 ÷ 36 = 11 ÷ 36 ≈ 0.31.
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