Search any question & find its solution
Question:
Answered & Verified by Expert
A fair coin is tossed 100 times. The probability of getting a head for even number of times is
Options:
Solution:
2031 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
We have $n=100$ and probability of getting head $=1 / 2$
Let $\mathrm{p}=1 / 2 \Rightarrow \mathrm{q}=1 / 2$
Probability of getting head even number of times $=$
$$
\begin{aligned}
& \mathrm{P}(\mathrm{X}=2)+(\mathrm{X}=4)+\ldots . .+(\mathrm{X}=100)] \\
& =\left[{ }^{100} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{98}+\ldots .+{ }^{100} \mathrm{C}_{100}\left(\frac{1}{2}\right)^{100}\left(\frac{1}{2}\right)^{\circ}\right] \\
& =\left(\frac{1}{2}\right)^{100}\left[{ }^{100} \mathrm{C}_2+{ }^{100} \mathrm{C}_4+\ldots .+{ }^{100} \mathrm{C}_{100}\right] \\
& =\left(\frac{1}{2}\right)^{100}\left[2^{100-1}\right]=\frac{1}{(2)^{100}} \times(2)^{99}=\frac{1}{2}
\end{aligned}
$$
Let $\mathrm{p}=1 / 2 \Rightarrow \mathrm{q}=1 / 2$
Probability of getting head even number of times $=$
$$
\begin{aligned}
& \mathrm{P}(\mathrm{X}=2)+(\mathrm{X}=4)+\ldots . .+(\mathrm{X}=100)] \\
& =\left[{ }^{100} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{98}+\ldots .+{ }^{100} \mathrm{C}_{100}\left(\frac{1}{2}\right)^{100}\left(\frac{1}{2}\right)^{\circ}\right] \\
& =\left(\frac{1}{2}\right)^{100}\left[{ }^{100} \mathrm{C}_2+{ }^{100} \mathrm{C}_4+\ldots .+{ }^{100} \mathrm{C}_{100}\right] \\
& =\left(\frac{1}{2}\right)^{100}\left[2^{100-1}\right]=\frac{1}{(2)^{100}} \times(2)^{99}=\frac{1}{2}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.