Question

Team A has the probability $\frac{2}{5}$ of winning, whenever it plays. If A plays 4 games, the probability that A wins more than half of the games is

• A. $\frac{22}{125}$
• B. $\frac{112}{625}$
• C. $\frac{116}{625}$
• D. $\frac{110}{625}$

Answer 1

The correct option is B $\frac{112}{625}$

Team $A$ has the probability of winning $=p=\frac{2}{5}$
A plays $4$ games $=n=4$
The probability that $A$ wins more than half of the games ${=}^4{C}_3(\frac{2}{5}{)}^3{(1-\frac{2}{5})}^{4-3}{+}^4{C}_4{\left(\frac{2}{5}\right)}^4{(1-\frac{2}{5})}^{4-4}$
$=\frac{96}{625}+\frac{16}{625}$

$=\frac{112}{625}$