Question

One hundred identical coins, each with probability p of showing up heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is:

• A. $\frac{1}{2}$
• B. $\frac{49}{101}$
• C. $\frac{50}{101}$
• D. $\frac{51}{101}$

Answer 1

The correct option is D $\frac{51}{101}$

$P_1{=}^{100}{C}_{50}\left(p{)}^{50}\right(1-p{)}^{50}{P}_2{=}^{100}{C}_{51}\left(p{)}^{51}\right(1-p{)}^{49}.Given{P}_1=P_2{\therefore }^{100}{C}_{50}\left(p{)}^{50}\right(1-p{)}^{50}{=}^{100}{C}_{51}\left(p{)}^{51}\right(1-p{)}^{49}.\frac{1-p}{p}=\frac{\frac{\left(100\right)!}{\left(51\right)!\left(49\right)!}}{\frac{\left(100\right)!}{\left(50\right)!\left(50\right)!}}=\frac{50}{51}\therefore p=\frac{51}{101}$