Eventually tossed a coin many times. Part-II

We can get at least 6 heads by = 10C6 + 10C7 + 10C8 + 10C9 +10C10 = 386 Probability = 386/1024 = 193/512 a+b = 705

(10C10+10C9+10C8+10C7+10C6)/2^10=193/512

Because:

$\sum_{n=0}^{10} {10 \choose n} = 2^{10} \text{ and } {n \choose k} = {n \choose n-k}$

the probability is:

$\frac{2^{10} - {10 \choose 5}}{2^{11}} = \frac{1024-252}{2^{11}} = \frac{772}{2048} = \frac{193}{512}$

and hence the answer is $193 + 512 = 705$ .

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Note: By using the two stated equalities we can limit the calculations to ${10 \choose 5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot2 \cdot 1}$ which cancels to $2 \cdot 9 \cdot 2 \cdot 7 = 252$ and reducing $\frac{772}{2048}$ to $\frac{386}{1024}$ and finaly $\frac{193}{512}$ .