Probability paradox (no it's not but can someone explain) In this little problem:
Amédée, Barnabé, Charles shoot a bird; if the probability of success is for Amédée: 70%, Barnabé:
50%, Charles: 90%, what is the probability that the bird will be hit?
I responded this way: a probability of 70% means that in 70 out of 100 cases the target is hit, so let's suppose that Amédée, Barnabé, Charles shoot 100 shot each one, the numbers of times they hit the target will be (70+50+90)/300 = 70% .
But the solution the book says is: Let us rather consider the complementary event: the bird is not touched if it is not touched either by Amédée, neither by Barnabas nor by Charles. This event has the probability: (1−0.7)·(1−0.5)·(1−0.9) =0.015.
The probability that the bird is hit is therefore: 1−0.015 = 0.985
which is 98.5%. Can someone explain where am I wrong (I know I am)?
 A: You calculated the probability that the  bird will get shot given that Amedee, Barnebe, and Charles are drawn at random and then ONE of them shoots. You were instead asked to calculate the probability that the [poor] bird gets shot given that each of them gets one shot.
It may help to ponder the following. If someone flips a quarter i.e., a fair coin $100$ times, then he can expect to get $50$ heads. Each time he flips he will get a head with a probability of $50\%$. Compare this to the probability that he gets heads at least once in his $100$ flips. The probability that he did not land a head even once in his $100$ flips, is $(1/2)^{100}$, which is far far less than $50/%$ and is fact fact far less than one-in-a-trillion. So the probability that he gets heads at least once is $1$ minus the less-than-one-in-a-trillion odds he gets no heads.
A: You computed the average $\frac{.7+.5+.9}{3}=\frac{2.1}{3}=.7$. This tells us that if these three shooters each perform an experiment where they shoot at a bird, their average performance is $.7$. In our experiment, three shooters are shooting at the same bird.
If $A$ goes first there is a $.7$ probability that the bird is shot and a $.3$ probability that it isn't.
If $A$ misses, there is a $.5$ probability that $B$ hits the bird and a $.5$ probability that $B$ misses.
If $B$ misses, there is a $.9$ probability that $C$ hits the bird and a $.1$ probability that the bird survives.
Let $A'$ be the probability that $A$ hits the bird. Define $B'$ and $C'$ similarly.
Then the probability that the bird gets hit will be $$P(A')+P(B'|\lnot A')P(\lnot A')+ P(C'|\lnot B',\lnot A')=.7 +.5(.3)+.9(.3)(.5)=.7+.15+.135=.985$$
