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If a fair coin is tossed 5 times, the probability that heads does not occur two or more times in a row is
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2347 Upvotes
Verified Answer
The correct answer is:
$\frac{13}{2^{5}}$
Case (1): All tail $\left(\frac{1}{2}\right)^{5}$
Case (2): $4 \mathrm{~T}, 1 \mathrm{H} \quad{ }^{5} \mathrm{C}_{4}\left(\frac{1}{2}\right)^{4} \cdot\left(\frac{1}{2}\right)^{5}=\frac{5}{2^{5}}$
Case (3): $\times \mathrm{T} \times \mathrm{T} \times \mathrm{T} \times$
$$
\left(\frac{1}{2}\right)^{3} \times{ }^{4} \mathrm{C}_{2} \times\left(\frac{1}{2}\right)^{2}=\frac{1}{2^{5}} \times 6=\frac{6}{2^{5}}
$$
Case (4): $\times \mathrm{T} \times \mathrm{T} \times$
$$
\left(\frac{1}{2}\right)^{2} \times\left(\frac{1}{2}\right)^{3}=\frac{1}{2^{5}}
$$
overall $\frac{13}{2^{5}}$
Case (2): $4 \mathrm{~T}, 1 \mathrm{H} \quad{ }^{5} \mathrm{C}_{4}\left(\frac{1}{2}\right)^{4} \cdot\left(\frac{1}{2}\right)^{5}=\frac{5}{2^{5}}$
Case (3): $\times \mathrm{T} \times \mathrm{T} \times \mathrm{T} \times$
$$
\left(\frac{1}{2}\right)^{3} \times{ }^{4} \mathrm{C}_{2} \times\left(\frac{1}{2}\right)^{2}=\frac{1}{2^{5}} \times 6=\frac{6}{2^{5}}
$$
Case (4): $\times \mathrm{T} \times \mathrm{T} \times$
$$
\left(\frac{1}{2}\right)^{2} \times\left(\frac{1}{2}\right)^{3}=\frac{1}{2^{5}}
$$
overall $\frac{13}{2^{5}}$
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