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A fair coin is tossed 100 times. The probability of getting tails an odd number of times is
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Probability of getting a tail in a single toss
$p=\frac{1}{2}$ and not getting tail, $q=\frac{1}{2}$.
Using Binomial distribution,
$\therefore$ Required probability
$\begin{aligned} & =P(X=1)+P(X=3)+P(X=5)+\ldots \\ & +P(X=99) \\ & ={ }^{100} C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{99}+{ }^{100} C_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{97} \\ & +{ }^{100} C_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{95}+\ldots .+{ }^{100} C_{99}\left(\frac{1}{2}\right)^{99}\left(\frac{1}{2}\right) \\ & =\left(\frac{1}{2}\right)^{100}\left({ }^{100} C_1+{ }^{100} C_3+{ }^{100} C_5+\ldots+{ }^{100} C_{99}\right) \\ & =\frac{1}{2^{200}}\left(2^{99}\right) \\ & =\frac{1}{2} \\ & \end{aligned}$
$p=\frac{1}{2}$ and not getting tail, $q=\frac{1}{2}$.
Using Binomial distribution,
$\therefore$ Required probability
$\begin{aligned} & =P(X=1)+P(X=3)+P(X=5)+\ldots \\ & +P(X=99) \\ & ={ }^{100} C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{99}+{ }^{100} C_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{97} \\ & +{ }^{100} C_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{95}+\ldots .+{ }^{100} C_{99}\left(\frac{1}{2}\right)^{99}\left(\frac{1}{2}\right) \\ & =\left(\frac{1}{2}\right)^{100}\left({ }^{100} C_1+{ }^{100} C_3+{ }^{100} C_5+\ldots+{ }^{100} C_{99}\right) \\ & =\frac{1}{2^{200}}\left(2^{99}\right) \\ & =\frac{1}{2} \\ & \end{aligned}$
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