If a fair coin is tossed 10 times the probability of atleast 6 heads is: Option 1: 105/512 Option 2: 53/128 Option 3: 53/64 Option 4: 193/512

Answer: 193/512. Solution: P(X ≥ 6) = _(k=6)^10 10k(1/2)^10. Coefficients: 106=210, 107=120, 108=45, 109=10, 1010=1. Sum = 386. So P = 386/1024 = 193/512.

Algebra

Probability

Medium

core

If a fair coin is tossed 10 times the probability of atleast 6 heads is:

$\frac{105}{512}$

$\frac{53}{128}$

$\frac{53}{64}$

$\frac{193}{512}$

✅ Correct Option: 4

$P(X \ge 6) = \sum_{k=6}^{10} \binom{10}{k}(1/2)^{10}$ . Coefficients: $\binom{10}{6}=210$ , $\binom{10}{7}=120$ , $\binom{10}{8}=45$ , $\binom{10}{9}=10$ , $\binom{10}{10}=1$ . Sum $= 386$ . So $P = 386/1024 = 193/512$ .

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