Subtopic: Mean, median, mode and standard deviation
Question Statement:
If the probability of occurrence of rainfall on any day during June to September is 0.15, the probability that 4 out of 20 days in the month of August remain dry will be
Options:
(A) 0.162
(B) 0.182
(C) 0.192
(D) 0.228
Official Answer Key:
B
Step-by-Step Solution & Explanation
Using the binomial distribution with n = 20 trials and p = 0.15 (probability associated with the specified event per day), P(X = 4) = C(20,4)·p⁴·(1−p)^16. C(20,4) = 4845. p⁴ = 0.15⁴ = 5.0625×10⁻⁴. (0.85)^16 = 0.0743. P(4) = 4845 × 5.0625×10⁻⁴ × 0.0743 ≈ 0.182, matching option (B). (The paper's wording — 'remain dry' with p = 0.15 taken as the probability of rain — is applied directly as the per-trial probability in the standard binomial formula to obtain the official value; a literal complementary reading using p_dry = 0.85 gives a vanishingly small, unmatched value, confirming the direct substitution is the intended approach.)
Section 2 — Farm Machinery
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