A dice is thrown twice, the probability of occurence of 5 at least once is

A dice is thrown twice. Finding the probability of getting 5 at least once.

"At least once" means getting 5 on the first throw, or the second throw, or both throws.

Using the complement rule:

$P(\text{at least one 5}) = 1 - P(\text{no 5 in both throws})$

When a dice is thrown once:

Total outcomes $= 6$ (numbers 1, 2, 3, 4, 5, 6)

Outcomes without 5 $= 5$ (numbers 1, 2, 3, 4, 6)

$P(\text{not getting 5}) = \frac{5}{6}$

Since the throws are independent:

$P(\text{no 5 in both}) = P(\text{no 5 in 1st throw}) \times P(\text{no 5 in 2nd throw})$

$P(\text{no 5 in both}) = \frac{5}{6} \times \frac{5}{6}$

$P(\text{no 5 in both}) = \frac{25}{36}$

$P(\text{at least one 5}) = 1 - P(\text{no 5 in both throws})$

$P(\text{at least one 5}) = 1 - \frac{25}{36}$

$P(\text{at least one 5}) = \frac{36}{36} - \frac{25}{36}$

$P(\text{at least one 5}) = \frac{11}{36}$

Therefore, the probability of getting 5 at least once is $\frac{11}{36}$.