A fair coin is tossed 6 times in succession. The probability of obtaining at least 5 heads is:

(A) $\frac{7}{64}$

(B) $\frac{6}{64}$

(C) $\frac{1}{64}$

(D) $\frac{11}{64}$

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles For $n = 6$ independent Bernoulli trials with success probability $p = 1/2$: $$P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} = \binom{6}{k} \left(\frac{1}{2}\right)^6 = \frac{\binom{6}{k}}{64}$$ "At least 5 heads" corresponds to the union of two mutually exclusive events: $k = 5$ or $k = 6$. ### 2. Analytical Formulation & Calculation $$P(X \ge 5) = P(X = 5) + P(X = 6) = \frac{\binom{6}{5} + \binom{6}{6}}{64}$$ $$\binom{6}{5} = 6, \quad \binom{6}{6} = 1$$ $$P(X \ge 5) = \frac{6 + 1}{64} = \frac{7}{64}$$ ### 3. Distractor Forensics - **Option A (7/64):** Correct sum of $k=5$ and $k=6$. - **Option B (6/64):** Only considers exactly 5 heads, neglecting all 6 heads. - **Option C (1/64):** Only considers exactly 6 heads. ### 4. Exam Trap & Standard Reference - **Trap Alert:** "At least" $k$ always includes all values from $k$ up to $n$. - **Standard Reference:** Aggarwal, R.S., *Quantitative Aptitude*.

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