Probability of an event given that one outcome didn't occur I have a homework question that I'm struggling with. It concerns conditional probability, but I can't figure out how to work in the condition. The problem is:
A dart thrower has a 75% chance of hitting the center. She throws three darts, then tells us she didn't hit the center 3 times. What is the probability she hit the center once?
I got the P(hitting the center once) = 1-(3*.75 = 2.25*(1-.75) = .5625^3 = .178) = .822
and I think the equation is P(hitting the center once/1-hitting 3 bullseyes) but the answer I'm getting doesn't seem to be correct. Any help would be great!
 A: Hints:
Let $H$ denotes the number of center hits. 
To be found is $\Pr\left(H=1\mid H<3\right)$
and we have the relation:
$$\Pr\left(H=1\mid H<3\right)\Pr\left(H<3\right)=\Pr\left(H=1\wedge H<3\right)=\Pr\left(H=1\right)$$
So if you can find $\Pr\left(H<3\right)$ and $\Pr\left(H=1\right)$
then you can also find $\Pr\left(H=1\mid H<3\right)$.
Also observe that $\Pr\left(H<3\right)=1-\Pr\left(H=3\right)$. 
A: Good question. I think the condition is the number of throws which doesn't matter because we're given $P(3 throws) = 1$. Since events with probability 0 or 1 are independent of any other event, including themselves, we can treat probabilities conditioned on them as unconditional probabilities.
$P($hit once | 1 throw) = $.75$
$P($hit a times | a throws) = $.75^a$
$P($hit 0 times | a throws) = $.75^0 (1-.75)^a$
$P($hit 1 time | a throws) = ? (for $a > 1$)
Consider rearranging
$$H\underbrace{MMM...MM}_{a-1}$$
$P($hit 1 time | a throws) = $\binom{a}{1}0.75^1(1-.75)^{a-1}$
$P($hit 1 time | 3 throws) = $\binom{3}{1}0.75^1(1-.75)^{3-1}$
The last one is 14%.
btw, $$\binom{3}{0}0.75^0(1-.75)^{3-0} < 0.02$$
I am 98% sure she is being too humble.
