Question

A fair coin is tossed $100$ times the probability of getting tails an odd number of time

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $0$
• D. $1$

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for the correct option:

Step1. Define Events:

Number of times coin tossed $\left(\mathrm{n}\right)=100$

Event of getting tails an odd number of time $=\mathrm{X}$

Probability of getting a tail after tossing a coin $\left(\mathrm{p}\right)=\frac{1}{2}$

Step2. Calculate the probability of getting tails an odd number of time

Probability of getting a tail after tossing a coin $\mathrm{X}=1,3,5,..............99$

$\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=3\right)+\mathrm{P}\left(\mathrm{X}=5\right)+...........+\mathrm{P}\left(\mathrm{X}=99\right)$

$$\begin{gathered} ={}^{100}{}^{\mathrm{C}}1\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^{100-1}+{}^{100}{}^{\mathrm{C}}3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{100-3}+....+{}^{100}{}^{\mathrm{C}}99{\left(\frac{1}{2}\right)}^{99}{\left(\frac{1}{2}\right)}^{100-99} \\ \mathbf{[}\mathbf{\because }\mathbf{Sum}\mathbf{}\mathbf{of}\mathbf{}\mathbf{odd}\mathbf{}\mathbf{terms}\mathbf{}\mathbf{}\mathbf{expansion}\mathbf{}{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{n}}\mathbf{=}{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{1}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{b}\mathbf{+}{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{3}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{3}}{\mathbf{b}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{n}\mathbf{-}\mathbf{1}{\mathbf{ab}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{]} \end{gathered}$$

$$\begin{gathered} ={}^{100}{}^{\mathrm{C}}1{\left(\frac{1}{2}\right)}^{100}+{}^{100}{}^{\mathrm{C}}3{\left(\frac{1}{2}\right)}^{100}+....+{}^{100}{}^{\mathrm{C}}99{\left(\frac{1}{2}\right)}^{100} \\ =\left[{}^{100}{}^{\mathrm{C}}1+{}^{100}{}^{\mathrm{C}}3+....+{}^{100}{}^{\mathrm{C}}99\right]{\left(\frac{1}{2}\right)}^{100} \\ =2^{99}{\left(\frac{1}{2}\right)}^{100}\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{1}\mathbf{+}{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{3}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{=}{\mathbf{2}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{]} \end{gathered}$$

$=\frac{\mathbf{1}}{\mathbf{2}}$

Hence, Option(A) is the correct answer.