How likely is to win at least 50 times out of 120 rolls? It is played on a 6-sided dice where you would win on 5-6 but lose on 1-4 so a 1/3 chance of winning. I know the probability of winning is 0.033 or 3.3% but I don't know how to arrive at that answer.
 A: The Bernoulli distribution gives the chance of a certain number of successes in a number of trials.  If you have $n=120$ trials, the chance of exactly $50$ successes is ${120 \choose 50}(1/3)^{50}(2/3)^{(120-50)}$  You need to sum over all the numbers of successes starting at $50$.  They will decrease rather rapidly, so sum them until you get tired and you will be close.
A: The random variable that you are considering is binomial distributed: $X\sim \mbox{Bin}(\mu,\sigma)$ where
$$\mu=n\cdot p=120\cdot(1/3)=40, \ 
\sigma=\sqrt{n\cdot p\cdot (1-p)}=\sqrt{120\cdot 1/3\cdot 2/3}\approx 5.164.$$
The exact probability that you have to determine is
$$P(X \geq 50)=\sum_{k=50}^{120}{120 \choose k}(1/3)^{k}(2/3)^{(120-k)}.$$
It is not easy to compute this formula with a pocket calculator, but since $n=120$, it can be approximated by the standard normal function $\Phi(z)$:
$$P(X \leq 49)\approx\Phi\left( \frac{49.5-\mu}{\sigma} \right)=\Phi(1.84)\approx 0.9671$$
The value for $\Phi(z)$ can be found in a table of a standard normal distribution.
Thus $P(X \geq 50)=1-P(X\leq 49)\approx 0.0329$.
