Five dice are thrown simultaneously.If the occurrence of an even number in a single dice is considered a success, then the probability of at most 3 successes is

$\frac{13}{16}$

The correct answer is Option (2) → $\frac{13}{16}$ **

Success probability on one die = $\frac{1}{2}$

Number of trials = $5$

$P(X\le 3)=\sum_{k=0}^{3}\frac{5!}{k!(5-k)!}\left(\frac{1}{2}\right)^5$

Compute terms:

For $k=0$: $\frac{5!}{0!5!}\left(\frac{1}{2}\right)^5 = 1\cdot\frac{1}{32}=\frac{1}{32}$

For $k=1$: $\frac{5!}{1!4!}\left(\frac{1}{2}\right)^5 = 5\cdot\frac{1}{32}=\frac{5}{32}$

For $k=2$: $\frac{5!}{2!3!}\left(\frac{1}{2}\right)^5 = 10\cdot\frac{1}{32}=\frac{10}{32}$

For $k=3$: $\frac{5!}{3!2!}\left(\frac{1}{2}\right)^5 = 10\cdot\frac{1}{32}=\frac{10}{32}$

Total $=\frac{1+5+10+10}{32}=\frac{26}{32}=\frac{13}{16}$

The probability of at most 3 successes is $\frac{13}{16}$.