Probability Distributions — Practice Quiz
Tamil Nadu Board · Class 12 · Mathematics
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Quick Quiz: Probability Distributions
0/4Tap an answer to check it instantly. No sign-up needed for these 4.
A fair coin is tossed twice. If X denotes the number of heads, what is P(X = 1)?
For a probability mass function f(x), which of the following conditions MUST be satisfied?
A random variable X has the probability mass function: f(1) = k, f(2) = 2k, f(3) = 3k. Find the value of k.
The cumulative distribution function F(x) is defined as:
Sample Questions
Given PMF: f(1) = 1/4, f(2) = 1/2, f(3) = 1/4. Find F(2), the CDF at x = 2.
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3/4
Step 1: F(2) = P(X ≤ 2) means we add all probabilities for x ≤ 2. Step 2: F(2) = f(1) + f(2) = 1/4 + 1/2. Step 3: = 1/4 + 2/4 = 3/4. Step 4: Note: F(3) = f(1)+f(2)+f(3) = 3/4 + 1/4 = 1, which confirms our total sums to 1. The answer is 3/4.
If F(x) is the CDF of a discrete random variable, what is P(X > x) equal to?
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1 - F(x)
Step 1: The total probability for all outcomes = 1. Step 2: P(X ≤ x) + P(X > x) = 1. Step 3: Since P(X ≤ x) = F(x), we get P(X > x) = 1 - F(x). Step 4: This is a standard complement rule applied to CDF. Example: if F(3) = 0.7, then P(X > 3) = 1 - 0.7 = 0.3.
For a continuous random variable X with PDF f(x), what is P(X = 5)?
A PDF f(x) = k for 2 ≤ x ≤ 6 and 0 otherwise. Find the value of k.
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1/4
Step 1: For a valid PDF, integral from -∞ to ∞ of f(x)dx = 1. Step 2: Since f(x) = k only for 2 ≤ x ≤ 6, we integrate k from 2 to 6. Step 3: k × (6 - 2) = 1 → k × 4 = 1. Step 4: k = 1/4. Verification: integral from 2 to 6 of (1/4)dx = (1/4)(4) = 1. ✓
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