Throw a dice 4 times. What is the probability `6` be up at-least one time? First time I approach a probability question (:




  Throw a dice 4 times. What is the probability 6 be up at-least one time?


Intuitively, I would say: $\frac{1}{6}\times4$.

I would explain as: If you throw one time, probability is $\frac{1}{6}$.

If you do it 4 times, then multiply by 4.

According to the answers I'm wrong.

Can you explain please? thanks in advance.
Expanding the comment of Vladimir Vargas about binomial distribution you can, alternatively, use the complementary probability to take zero 6's in the four throws:

$$P(X\ge 1)=1-P(X=0)=1-\binom{4}{0}\left(\frac16\right)^0\left(1-\frac16\right)^4=1-\left(\frac56\right)^4=\frac{671}{1296}\approx52\%$$
Consider the complement problem, there is a 5/6 probability of not rolling a six
for any given die, and since the four dice are independent, the probability of not rolling a six is (5/6)^4 = 5^4/6^4 = 625/1296. The probability of rolling at least one six is therefore
1 − 625/1296 = 671/1296 ≈ .517.