Question

In a set of games it is $3$ to $5$ in favour of the winner of the previous game.. Then the probability that a person who has won the first game shall win at least $2$ out of the next $5$ games is ?

• A. $\frac{3^4}{3^5}$
• B. ${\left(\frac{2}{5}\right)}^5$
• C. ${\left(\frac{3}{5}\right)}^5$
• D. ${\left(\frac{3}{5}\right)}^3{\left(\frac{2}{5}\right)}^2$

Answer 1

Answer: (A)

$\frac{3^4}{3^5}$

$p\left(w\right)=\frac{3}{5}p\left(l\right)=\frac{2}{5}$

first game shall write at $\left(B\right)/4$ game

$\Rightarrow \left(\frac{3}{5}\right)\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)\left(\frac{3}{5}\right){\left(\frac{2}{5}\right)}^1+\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)$

$\Rightarrow {\left(\frac{3}{5}\right)}^4[-\frac{2}{5}+\frac{3}{5}]$

$={\left(\frac{3}{5}\right)}^4\left(\frac{1}{5}\right)=\frac{3^4}{3^5}$