A coin is tossed $n$ times. The probability that head will turn up an odd number of times, is

\frac{1}{2}

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\frac{n + 1}{2 n}

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\frac{n - 1}{2 n}

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\frac{2^{n - 1} - 1}{2^{n}}

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Solution

The correct option is B $\frac{1}{2}$
Let $X$ denote the number of heads in $n$ trails. Then,
$P (X = r) =^{n} C_{r} {(\frac{1}{2})}^{r} {(\frac{1}{2})}^{n - r} =^{n} C_{r} {(\frac{1}{2})}^{n}$
Thus, the required probability
$= P (X = 1) + P (x = 3) + P (X = 5) + . . .$
$=^{n} C_{1} {(\frac{1}{2})}^{n} +^{n} C_{3} {(\frac{1}{2})}^{n} +^{n} C_{5} {(\frac{1}{2})}^{n} + . . .$
$= {(\frac{1}{2})}^{n} (^{n} C_{1} +^{n} C_{3} +^{n} C_{5} + . . .) = \frac{1}{2^{n}} (2^{n - 1}) = \frac{1}{2}$

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