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A coin is tossed 5 times. The probability that tail appears an odd number of times, is
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Let $\mathrm{p}$ denote the probability of getting tail in a single of a coin.
$\therefore \mathrm{p}=\frac{1}{2} \Rightarrow \mathrm{q}=\frac{1}{2} \& \mathrm{n}=5$
Let $X$ denot no. of tails in 5 tosses of coin. $\begin{aligned} \therefore \text { Required probability } &=\mathrm{P}(\mathrm{x}=1)+\mathrm{P}(\mathrm{x}=3)+\mathrm{P}(\mathrm{x}=5) \\ &=\frac{1}{2^{5}}\left[{ }^{5} \mathrm{C}_{1}+{ }^{5} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{5}\right] \\ &=\frac{1}{2^{5}}[5+10+1]=\frac{16}{32}=\frac{1}{2} \end{aligned}$
$\therefore \mathrm{p}=\frac{1}{2} \Rightarrow \mathrm{q}=\frac{1}{2} \& \mathrm{n}=5$
Let $X$ denot no. of tails in 5 tosses of coin. $\begin{aligned} \therefore \text { Required probability } &=\mathrm{P}(\mathrm{x}=1)+\mathrm{P}(\mathrm{x}=3)+\mathrm{P}(\mathrm{x}=5) \\ &=\frac{1}{2^{5}}\left[{ }^{5} \mathrm{C}_{1}+{ }^{5} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{5}\right] \\ &=\frac{1}{2^{5}}[5+10+1]=\frac{16}{32}=\frac{1}{2} \end{aligned}$
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