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A coin is tossed three times. What is the probability of getting head and tail alternately?
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The correct answer is:
$\frac{1}{4}$
Coin is tossed three times i.e. total outcomes $=2^{3}=8$ $[(\mathrm{H}, \mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{H}, \mathrm{T}),(\mathrm{H}, \mathrm{T}, \mathrm{H}),(\mathrm{H}, \mathrm{T}, \mathrm{T}),(\mathrm{T}, \mathrm{H}, \mathrm{H}),$,
$(\mathrm{T}, \mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{T}, \mathrm{T})]$
Alternate head and tail are coming two times only.
Thus prob. of getting head and tail alternately $=\frac{2}{8}=\frac{1}{4}$
$(\mathrm{T}, \mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{T}, \mathrm{T})]$
Alternate head and tail are coming two times only.
Thus prob. of getting head and tail alternately $=\frac{2}{8}=\frac{1}{4}$
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