I am stuck on the following question and am not sure where to go from what I have currently done.
Donald Trump has gone moose hunting. When he fires his gun at a moose he scores a hit with a probability of 0.8. On a typical hunt, he will fire his gun 30 times. Let X denote the number of shots that actually hit a moose.
I know i need to find out n, p and q. So n = 30, p = 0.8 and q = 0.2
i) What is the probability that at least 3 of the shots are misses?
For this do I need to find the probability that x=3, x=4, x=5 ... x=29, x=30?
that is all that I can think to do.
ii) What is the probability that more than 20 of the shots are hits?
for this one do I need to find the probability that x=20, x=21, ... x=30?
I know I have to put it into an equation, but I am not sure how to do that equation. any help will be great. thank you
Yes, specifically you want to find $P(X \geq 3) = \sum_{x = 3}^{30}P(X = x)$ where we can add because the events are disjoint. However, there are two problems. First, that's a lot of work. You instead might want to consider the complement and do $$P(X\geq 3) = 1- P(X<3).$$
Second, notice how $X$ was defined. It counts the number of hits (which are considered "successes"), but in part i., we want to count the number of misses. We consider the misses to be the "successes". So for clarity, define a new random variable $Y$ that follows the binomial distribution with $n = 30$ and $p = 0.2$, where $p$ is the probability of missing ("success") and compute $$P(Y\geq 3) = 1-P(Y<3).$$
Finally, recall that for a random variable $X$ that follows the binomial distribution with $n$, $p$, the probability that $X = k$ for $0\leq k\leq n$ is $$P(X= k) = \binom{n}{k}p^k(1-p)^{n-k}$$
where $\binom{n}{k}$ is the binomial coefficient and $p+q = 1$.