Probability of exactly $50$ heads One hundred identical coins, each with probability $p$ of showing up a head, are tossed. If $0 < p < 1$ and if
the probability of heads on exactly $50$ coins is equal to that of heads on exactly $51$ coins then the value of $p$,
is
$A. 1/2$
$B.49/101$
$C.50/101.$
$D. 51/101 . $
I don't know how to do it , I'm totally new to probability theory. Please give me some hints.
 A: The probability to get exactly $50$ heads on $100$ tossed coins is:
$$A= {100 \choose 50}p^{50}(1-p)^{100-50}.$$
The probability to get exactly $51$ heads on $100$ tossed coins is:
$$B= {100 \choose 51}p^{51}(1-p)^{100-51}.$$
Then:
$$A = B \Rightarrow {100 \choose 50}p^{50}(1-p)^{100-50} = {100 \choose 51}p^{51}(1-p)^{100-51} \\
\Rightarrow \frac{100!}{50!50!}p^{50}(1-p)^{50} = \frac{100!}{51!49!}p^{51}(1-p)^{49} \\
\Rightarrow \frac{1}{50!50!}(1-p) = \frac{1}{51!49!}p \\
\Rightarrow \frac{1}{50!50 \cdot 49!}(1-p) = \frac{1}{51 \cdot 50! 49!}p \\
\Rightarrow \frac{1}{50 }(1-p) = \frac{1}{51 }p \\
\Rightarrow p\left(\frac{1}{50}+\frac{1}{51}\right) = \frac{1}{50 }\\
\Rightarrow p\frac{50 + 51}{50 \cdot 51} = \frac{1}{50 }\\
\Rightarrow p\frac{101}{51} = 1\\
\Rightarrow p = \frac{51}{101}. $$
A: We have the equality
$$
{100\choose 50}p^{50}(1-p)^{50}={100\choose 51}p^{51}(1-p)^{49}
$$
and this equality gives us the value of $p$ (see the binomial distribution).
A: $n=100$ , so lets start $P(H)=p so P(T)=1-p$ now its nothing but binomial theorem  ${n\choose r}p^r.q^{n-r}$ where $p=successes,q=failures$. So $${100\choose 50}.p^{50}.(1-p)^{50}={100\choose 51}.p^{51}.(1-p)^{49}$$ solving you get $p=51/101$ hope its clear
A: Probability of $k$ successes in $n$ trials is given by the formula ${n \choose k} p^k (1-p)^{n-k}$ (binomial distribution), where $p$ is the probability of success.
We are given that:
$$
\frac{100!}{(50!)^2}p^{50}(1-p)^{50}=\frac{100!}{51!\cdot49!}p^{51}(1-p)^{49}
$$
It means that:
$$
\frac{51}{50}(1-p)=p \implies 50p=51-51p = \implies p =\frac{51}{101}
$$
