One hundred identical coins are thrown as each coin has the probability of head as p. Let X = number of coins showing heads then match the following conditions with p value.

A) P(X = 49) = P(X= 50)          1) $p=\frac{1}{2}$

B) P(X= 48) = P(X = 52)          2) $\frac{51}{101}$

C) P(X = r) = P(X = n – r)          3) $\frac{50}{101}$

                                                    4) $\frac{57}{100}$

The correct matching is

Let X be the number of coins showing heads then $\mathrm{X}~\mathrm{B} (100, \mathrm{P})$

A)     Now $\mathrm{P}(\mathrm{X}=49)=\mathrm{P}(\mathrm{X}=50)$

         $$\begin{gathered} \Rightarrow {}^{100}{\mathrm{c}}_{49} {\mathrm{p}}^{49} {\mathrm{q}}^{51}={}^{100}{\mathrm{c}}_{50} {\mathrm{p}}^{50} {\mathrm{q}}^{50} \\ \Rightarrow {}^{100}{\mathrm{c}}_{49} \mathrm{q}={}^{100}{\mathrm{c}}_{50} \mathrm{P} \\ \Rightarrow \frac{\mathrm{q}}{\mathrm{p}}=\frac{{}^{100}{\mathrm{c}}_{50}}{{}^{100}{\mathrm{c}}_{49}} \\ =\frac{100!}{50! 50!}\times \frac{49! 51!}{100!} \\ \frac{\mathrm{q}}{\mathrm{p}}=\frac{49!\times 51\times 50!}{50\times 49!\times 50!} \end{gathered}$$

           $$\begin{gathered} \Rightarrow \frac{\mathrm{q}}{\mathrm{p}}=\frac{51}{50} \\ \Rightarrow 50(1-p)=51p \\ \Rightarrow 50-50p=51p \\ \Rightarrow 50=101p \\ \Rightarrow p=\frac{50}{101} \end{gathered}$$

B)     $\mathrm{P}(\mathrm{X}=48)=\mathrm{P}(\mathrm{X}=52)$

          $$\begin{gathered} \Rightarrow {}^{100}{\mathrm{c}}_{48} {\mathrm{p}}^{48} {\mathrm{q}}^{52}={}^{100}{\mathrm{c}}_{52} {\mathrm{p}}^{52} {\mathrm{q}}^{48} \\ \Rightarrow {}^{100}{\mathrm{c}}_{48} {\mathrm{q}}^4={}^{100}{\mathrm{c}}_{52} {\mathrm{p}}^4 \\ \Rightarrow \frac{{\mathrm{q}}^4}{{\mathrm{p}}^4}=\frac{100!}{52! 48!}\times \frac{48!\times 52!}{100!} \\ \Rightarrow {\left(\frac{\mathrm{q}}{\mathrm{p}}\right)}^4=1 \\ \Rightarrow \frac{\mathrm{q}}{\mathrm{p}}=1 \\ \Rightarrow 1-\mathrm{p}=p \\ \Rightarrow 2p=1 \\ \Rightarrow p=\frac{1}{2} \end{gathered}$$

C)     $\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{P}(\mathrm{X}=\mathrm{n}-\mathrm{r})$

         $$\begin{gathered} \Rightarrow {}^{100}{\mathrm{c}}_{\mathrm{r}} {\mathrm{p}}^{\mathrm{r}} {\mathrm{q}}^{\mathrm{n}-\mathrm{r}}={}^{100}{\mathrm{c}}_{\mathrm{n}-\mathrm{r}} {\mathrm{p}}^{\mathrm{n}-\mathrm{r}} {\mathrm{q}}^{\mathrm{r}} \\ \Rightarrow \frac{{\mathrm{p}}^{\mathrm{r}} {\mathrm{q}}^{\mathrm{n}-\mathrm{r}}}{{\mathrm{p}}^{\mathrm{n}-\mathrm{r}} {\mathrm{q}}^{\mathrm{r}}}=\frac{{}^{100}{\mathrm{c}}_{\mathrm{n}-\mathrm{r}}}{{}^{100}{\mathrm{c}}_{\mathrm{r}}} \\ \Rightarrow {\left(\frac{\mathrm{q}}{\mathrm{p}}\right)}^{\mathrm{n}-2\mathrm{r}}=1 \end{gathered}$$

As it is independent of n for every 'r' so $\frac{\mathrm{q}}{\mathrm{p}}=1$

         $$\begin{gathered} \Rightarrow \mathrm{q}=\mathrm{p} \\ \Rightarrow 1-p=p \\ \Rightarrow 2p=1 \\ \Rightarrow p=\frac{1}{2} \end{gathered}$$