Question

A biased coin with probability p, $0

• A. 44564
• B. 44595
• C. 44597
• D. 44625

Answer 1

Answer: (A)

44564

Answer 2

Let $q=1-p .$ Since, head appears first time in an even throw 2 or 4 or $6 \ldots .$ $\therefore \frac{2}{5}=q p+q^{3} p+q^{5} p+\ldots$ $\Rightarrow \frac{2}{5}=\frac{q p}{1-q^{2}}$ $\Rightarrow \frac{2}{5}=\frac{(1-p) p}{1-(1-p)^{2}}$ $\Rightarrow \frac{2}{5}=\frac{1-p}{2-p}$ $\Rightarrow 4-2 p=5-5 p$ $\Rightarrow 3 p=1$ $\Rightarrow p=\frac{1}{3}$