conditional probabilities exercises

Question

I've been practicing some conditional probabilities and got stuck on a few of them

1. Find $P(B^{\prime}|A^{\prime})$, if $P(A) = a$, $P(B) = b$ and $P(A \cap B) = c$. ($a = 0.48$, $b = 0.33$, $c = 0.17$)
2. Urn contains: $k$ white balls, $m$ red balls and $n$ black balls. without returning the balls are taken out of the urn. what is the probability of getting a white ball earlier than getting a black ball ($k = 5$, $m = 2$, $n = 7$)
3. events $A$ and $B$ probabilities are $a$ and $b$ respectively, the probability for one of these events to happen is $c$. it is known that event $B$ has happened, what is the probability of event $A$ happening as well.

i have an idea how to solve #1, but have no clue on the rest of them. i am not as much interested in answers as i am in solutions, so any input will be appreciated

Answer 1

for #1

$P(A'|B')=P(B'∩A')/P(A')$

$P(B'∩A') = P(A')+P(B') - P(A'∪B')$

$P(A'∪B') =1-P(A∩B)$

$P(A∩B)=0.17, P(A')=0.52, P(B')=0.67$ as given in the exercise

then we walk backwards

$P(A'∪B') =1-P(A∩B)=0.83$

$P(B'∩A') = P(A')+P(B') - P(A'∪B') = 0.36$

$P(A'|B')=P(B'∩A')/P(A') = ~0.6923$

#2 $\dfrac{k}{k+n}$ is the correct formula to sum up the question. as we need k which are white ball number and we divide it by the number of influential balls in the urn, red balls won't change the answer in any way as if you pick up one you will just draw another ball afterwards, the game will not end because of red ones so they can be excused from the calculations

#3 is in the answer given below

Answer 2

I found question 3 easier so answering it first.

You are given $P(A) = a$ and $P(B) = b$ and probability for occurrence of any of the given events is c i.e. $P(A \cup B) = c$ and you are to find $P(A|B) =\:?$

So, $P(A|B) = \frac{P(A \cap B)}{P(B)}$

Here you are not given that A and B are independent events so $P(A\cap B) \ne P(A).P(B)$

So taking $P(A \cap B) = P(A) + P(B) - P(A \cup B) = a + b - c$

$\therefore P(A|B) = \frac{a+b-c}{b}$

Above is my opinion and it seems correct at least to me. Do let me know if you don't feel good about it.