A fair coin is tossed 100 times. The probability of getting tails 1, 3, ……., 49 times is

Let the probability of getting a tail in a single trial be p = 1/2. The number of trials be n = 100 and the number of trials in 100 trials be X.

We have,

$\mathrm{P}(\mathrm{X}=\mathrm{r}){=}^{100}{\mathrm{C}}_{\mathrm{n}}{\mathrm{p}}^{\mathrm{r}}{\mathrm{q}}^{\mathrm{n}-\mathrm{r}}$

${=}^{100}{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{r}}{\left(\frac{1}{2}\right)}^{100-\mathrm{r}}{=}^{100}{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{100}$

Now,

$\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=3)+\cdots +\mathrm{P}(\mathrm{X}=49)$

           $\begin{array}{l}{=}^{100}{\mathrm{C}}_1{\left(\frac{1}{2}\right)}^{100}{+}^{100}{\mathrm{C}}_3{\left(\frac{1}{2}\right)}^{100}+\cdots {+}^{100}{\mathrm{C}}_{49}{\left(\frac{1}{2}\right)}^{100}\\ =\left({ }^{100}{\mathrm{C}}_1{+}^{100}{\mathrm{C}}_3+\cdots {+}^{100}{\mathrm{C}}_{49}\right){\left(\frac{1}{2}\right)}^{100}\end{array}$

But ${ }^{100}{\mathrm{C}}_1{+}^{100}{\mathrm{C}}_3+\cdots {+}^{100}{\mathrm{C}}_{99}=2^{99}$

Also, ${ }^{100}{\mathrm{C}}_{99}{=}^{100}{\mathrm{C}}_1$

${ }^{100}{\mathrm{C}}_{97}{=}^{100}{\mathrm{C}}_3,{\dots }^{100}{\mathrm{C}}_{51}{=}^{100}{\mathrm{C}}_{49}$

Thus, $2\left({ }^{100}{\mathrm{C}}_1{+}^{100}{\mathrm{C}}_3+\cdots {+}^{100}{\mathrm{C}}_{49}\right)=2^{99}$

or ${ }^{100}{\mathrm{C}}_1{+}^{100}{\mathrm{C}}_3+\cdots {+}^{100}{\mathrm{C}}_{49}=2^{98}$

Therefore, probability of required event is

$\frac{2^{98}}{2^{100}}=\frac{1}{4}{2}^{98}/2^{100}=1/4$