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Probability & Distributions
Paper-II Statistics (Post-specific) · Statistics Syllabus
Useful for: SSC CGL
Overview
Probability theory and standard distributions (binomial, Poisson, normal) model random outcomes.
Key points
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B); P(A | B) = P(A ∩ B)/P(B).
- Binomial: mean np, variance npq (q = 1 − p).
- Poisson: mean = variance = λ; P(X = x) = e^(−λ) λˣ/x!.
- Normal: symmetric, mean = median = mode; about 68%, 95%, 99.7% within 1, 2, 3 SDs.
- Standard normal z = (x − μ)/σ.
Explanation
- Bayes’ theorem updates probabilities given new evidence.
- Binomial variance is always less than its mean.
- Poisson approximates binomial when n is large and p small (λ = np).
- Area under the whole normal curve = 1.
Step by step
- Identify the measure or test asked and its formula.
- Organise the data (table, totals) before computing.
- Interpret the result (sign, size, significance).
Examples
- Binomial with n = 10, p = 0.4: mean = 4, variance = 2.4.
- Poisson with λ = 2: P(X = 0) = e^(−2) ≈ 0.135.
- About 95% of a normal distribution lies within ±2σ of the mean.
Traps and advanced points
- Trap: Using npq as the mean.
- Trap: Thinking Poisson variance differs from its mean.
- Trap: Applying the 68–95–99.7 rule to skewed data.
Quick revision
- Binomial np, npq.
- Poisson mean = variance.
- 68–95–99.7.
Practice MCQs
The probability of getting a head when a fair coin is tossed once is
- 1
- 1/2
- 0
- 1/4
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Answer: (b) 1/2. Two equally likely outcomes.
The probability of getting a sum of 7 with two fair dice is
- 1/36
- 1/12
- 7/36
- 1/6
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Answer: (d) 1/6. 6 favourable outcomes out of 36.
For a binomial distribution with n = 10 and p = 0.5, the mean is
- 0.5
- 2.5
- 10
- 5
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Answer: (d) 5. Mean = np.
In a Poisson distribution, the mean and variance are
- equal
- always zero
- mean twice the variance
- unrelated
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Answer: (a) equal. Both equal λ.
About what percentage of values in a normal distribution lie within ±1 standard deviation of the mean?
- 50%
- 95%
- 99.7%
- 68%
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Answer: (d) 68%. 95% lie within ±2 SD.
For two mutually exclusive events A and B, P(A ∪ B) equals
- P(A) + P(B)
- P(A) × P(B)
- P(A) − P(B)
- 1
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Answer: (a) P(A) + P(B). They cannot occur together.
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