Bernoulli Distributions — Full Worked Solutions
Full Worked Solutions
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Question: A fair coin is tossed. Define X = 1 if heads, X = 0 if tails. Show this is a Bernoulli trial and state p.
Identifying the Bernoulli Trial
A Bernoulli trial requires: (1) exactly two outcomes, (2) a fixed probability of success, (3) one toss of the coin.
Here: “Heads” = success (X = 1), “Tails” = failure (X = 0). The coin is fair, so the probability of success is constant and known.
p = P(Heads) = 0.5
Distribution: P(X = 1) = 0.5, P(X = 0) = 0.5
This is X ~ Bernoulli(0.5).
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Question: For X ~ Bernoulli(0.3), find P(X = 1) and P(X = 0).
Applying the Bernoulli Distribution
For X ~ Bernoulli(p), the two probabilities are:
P(X = 1) = p = 0.3
P(X = 0) = 1 − p = 1 − 0.3 = 0.7
Check: P(X = 0) + P(X = 1) = 0.7 + 0.3 = 1 ✓
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Question: For X ~ Bernoulli(0.7), find E(X).
Using the Bernoulli Mean Formula
For any Bernoulli(p) random variable: E(X) = p
E(X) = p = 0.7
Derivation: E(X) = 1 × P(X=1) + 0 × P(X=0) = 1(0.7) + 0(0.3) = 0.7 ✓
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Question: For X ~ Bernoulli(0.7), calculate Var(X) using the formula Var(X) = p(1 − p).
Computing the Variance
Var(X) = p(1 − p)
= 0.7 × (1 − 0.7)
= 0.7 × 0.3
= 0.21
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Question: For X ~ Bernoulli(0.7), find SD(X).
Standard Deviation from Variance
From Q4: Var(X) = 0.21
SD(X) = √Var(X) = √0.21 ≈ 0.458
Alternatively: SD(X) = √(p(1 − p)) = √(0.7 × 0.3) = √0.21 ≈ 0.458
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Question: A randomly selected student passes a surprise quiz with probability 0.15. Define X = 1 if they pass and X = 0 if not. State the distribution and find E(X) and SD(X).
Define the Distribution
X = 1 (pass) with probability 0.15, X = 0 (fail) with probability 0.85.
X ~ Bernoulli(0.15), so p = 0.15.
Mean and Standard Deviation
E(X) = p = 0.15
Var(X) = p(1 − p) = 0.15 × 0.85 = 0.1275
SD(X) = √0.1275 ≈ 0.357
Interpretation: On average, 15% of students pass, with a standard deviation of about 0.357.
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Question: Compare X ~ Bernoulli(0.2) and Y ~ Bernoulli(0.8). Which has greater variance, and why?
Computing Variances
Var(X) = 0.2 × 0.8 = 0.16
Var(Y) = 0.8 × 0.2 = 0.16
Interpretation
The variances are equal. This is because p(1 − p) is symmetric: plugging in p gives the same result as plugging in (1 − p).
More intuitively: a trial that succeeds 20% of the time has the same uncertainty as one that succeeds 80% of the time — both are “far from certain” in the same way. Variance only depends on how far p is from the extremes 0 and 1.
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Question: X ~ Bernoulli(0.4). Find E(5X + 3) and Var(5X + 3).
Linear Transformation Rules
For constants a = 5 and b = 3:
E(aX + b) = aE(X) + b
Var(aX + b) = a²Var(X)
Calculations
E(X) = p = 0.4, Var(X) = 0.4 × 0.6 = 0.24
E(5X + 3) = 5(0.4) + 3 = 2 + 3 = 5
Var(5X + 3) = 5² × 0.24 = 25 × 0.24 = 6
Note: the constant +3 shifts the mean but does not affect variance.
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Question: Draw the probability distribution table for X ~ Bernoulli(0.6) and verify that ∑ P(X = x) = 1.
Distribution Table
For X ~ Bernoulli(0.6), p = 0.6:
x 0 1 P(X = x) 0.4 0.6 Verification: ∑ P(X = x) = 0.4 + 0.6 = 1 ✓
E(X) = 0(0.4) + 1(0.6) = 0.6 = p ✓
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Question: A factory produces bolts. Historically, 3% of bolts are defective. One bolt is randomly selected and inspected. Define an appropriate Bernoulli random variable, state its distribution, and find the mean and variance. Interpret E(X) in context.
Define the Random Variable
Let X = 1 if the selected bolt is defective, and X = 0 if it is not defective.
P(defective) = 0.03, so X ~ Bernoulli(0.03) with p = 0.03.
Distribution
P(X = 1) = 0.03 P(X = 0) = 0.97
Mean and Variance
E(X) = p = 0.03
Var(X) = p(1 − p) = 0.03 × 0.97 = 0.0291
Interpretation
E(X) = 0.03 means that, on average, 3% of bolts inspected are defective. If we repeatedly performed this experiment (selecting one bolt at a time), the long-run proportion of X = 1 outcomes would approach 3%.
SD(X) = √0.0291 ≈ 0.171, indicating the typical deviation from the mean on any single inspection.