I am working on some exercises related to random variables, since the subject is new to me, I feel a little insecure about my answers, I'll write the problem I've worked on to check if my solution is correct. I would appreciate if someone could take a look at the solution:
Problem
The probability of hitting a target is $\frac{1}{5}$. Ten shots are fired independently.
1) Find the probability of hitting the target at least twice.
2) Find the conditional probability of hitting at least twice knowing the target was hit at least once.
I define the random variable $X$ to be the number of times the target is hit. For 1), I am asked to calculate $P(X\geq 2)$, but $$P(X \geq 2)=1-P(X <2)=1-(\binom{10}{0}(\frac{4}{5})^{10}+\binom{10}{1}(\frac{1}{5})(\frac{4}{5})^9)$$
I have more doubts with part two: I want to calculate $P(X \geq 2| X\geq 1)$, so $$P(X \geq 2| X\geq 1)=\frac{P(X \geq 2 \cap X \geq 1)}{P(X\geq 1)}$$. Since $\{X \geq 2\} \subset \{X \geq 1\}$, we have $$P(X \geq 2| X\geq 1)=\frac{P(X \geq 2)}{P(X \geq 1)}$$
$P(X \geq 1)=1-P(X=0)=1-\binom{10}{0}(\frac{4}{5})^{10}$. By this last calculation and by 1), finally we have $P(X \geq 2)=\dfrac{1-(\binom{10}{0}(\frac{4}{5})^{10}+\binom{10}{1}(\frac{1}{5})(\frac{4}{5})^9)}{1-\binom{10}{0}(\frac{4}{5})^{10}}$
At a first glance, it seems to me that 2) makes sense since it is a bigger value than the probability in 1), but that doesn't necessarily mean the answer is correct. I wouldr really appreciate if someone could check my solution.
Best Answer
Your solution is correct. You may also give an alternative justification for your simplification in the second part by applying Bayes Theorem.
$P(X\geq 2 \vert X\geq 1) = \frac{P(X\geq 1 \vert X\geq 2)P(X\geq 2)}{P(X\geq 1)}$
Observing that $P(X\geq 1 \vert X\geq 2) = 1$ we obtain
$P(X\geq 2 \vert X\geq 1) = \frac{P(X\geq 2)}{P(X\geq 1)}$
The details of which you have already worked out.