The probability of a person hitting a target is 1/4.
1.) If they fire five times, what is the probability of hitting the target exactly four times?
2.) What is the probability of hitting the target at least twice?
My attempts:
1.) With the formula
where n = 5, k = 4, p = 1/4, I got 0.0146
2.) I did the same thing but this time k = 2 and got 0.2637. But this is wrong. So my question is how do I find the probability of this.
The kick? "At least." The PMF (probability mass function) for a binomial distribution with parameters $n, p,$ is
$$ P(X = x) = \binom{n}{k} p^k (1-p)^{n-k}. $$
Now, to get the right answer, you need
$$ P(X \geq 2) = \sum_{j=2}^{5} \binom{5}{j} \left( \frac{1}{4} \right)^j \left( \frac{3}{4} \right) ^{5-j}. $$
Can you take it from here, champ? You got it!
(Of course the quicker kill is $P(X \geq 2) = 1 - P(X \leq 1)$, but either way works.)