Let us have a urn with 10 balls, the balls are either white or black ( we don't know in which proportions ).We extract 4 balls without reintroducing them back and after each extraction the ball is white, meaning that the 4 balls extracted were all white.I need to find the probability that the urn contains only white balls.
My solution ( but I'm not so sure it's correct ):
we have a white balls and 10-a black balls
I started off by naming the events:
B = " urn contains only white balls "
A(i) = " white ball appears i times in n extractions" for our case n=4 and i=4
and I was thought I should use bayes to find my probability:
$$ P(B/A_4)=\frac{P(B)*P(A_4/B)}{P(A_4)} $$
$P(B/A_4) $- the probability of all the balls in the urn being white when we know that in 4 extractions all the balls were white
$P(A_4/B) $- probability that in 4 extractions the balls will be white knowing that the urn contains only white balls ( which is equal to 1)
$P(A_4) $-the probability that we will have 4 white balls in 4 extractions
$P(B)$-the probability of all the balls being white
And then started my calculus (at least tried):
For $P(B)$ I thought that if we extracted 1 ball from the urn the probability of it being white would be $\frac{a}{10}$ so that for 10 balls it will be $\frac{a}{10}^{10}$ (and here is my question should I have used the hypergeometric distribution here? I mean to calculate the probability that in 10 extractions (we don't introduce them back) all the balls will be white:$\frac{{{a}\choose{10}}*{{10-a\choose{0}}}}{10\choose{10}} $ I not sure what should I do here and I would appreciate some help.
For$ P(A_4)$ I used the hypergeometric distribution $ \frac{{{a}\choose{4}}*{{10-a\choose{0}}}}{10\choose{4}} $
I would really appreciate some help with this problem and don't hold back on criticism if you see I did some mistakes (I'm new to probabilities) and thank you in advance.