Question

A certain coin lands head with probability $p$. Let $Q$ denote the probability that when the coin is tossed four times the number of heads obtained is even. Then

• A. there is no value of $p$, if $Q =\frac{1}{4}$
• B. there is exactly one value of $P$ if $Q =\frac{3}{4}$
• C. there are exactly two values of $p$ if $Q =\frac{3}{5}$
• D. there are exactly four values of $p$ if $Q =\frac{4}{5}$

Answer 1

Answer: (A), (C)

$ P (0 H$ or $2 H$ or $4 H )$
$Q=p^4+6 p^2(1-p)^2+(1-p)^4=8 p^4-16 p^3+12 p^2-4 p+1=\frac{(2 p-1)^4+1}{2}$. Now interpret