conditional probabilites - pairs of balls

Question

Been struggling with an exercise for quite some time as i can't get the correct answer, so decided to get some help.

there are 8 black and 8 white balls in the urn. we randomly pull 4 balls in a row. let’s mark the events:

• A1 - (1-2 ball pair contains at least 1 white ball)
• A2 - (2-3 ball pair contains at least 1 black ball)
• A3 - (3-4 ball pair are both black)

simplify events $A = \neg(\neg A_1∪A_2)∩A_3$ and find probability $P(A)$.

what I did is first of all simplify the above event:

$A1∩\neg A2∩A3$

and try to calculate the events

$A1= \frac{8}{16}*\frac{7}{15}+\frac{8}{16}*\frac{8}{15}+\frac{8}{16}*\frac{8}{15}$ which is the same as $1-\frac{8}{16}*\frac{7}{15}$

so I did the same with the rest

$A2 = 1-\frac{8}{16}*\frac{7}{15}$

$A3 = \frac{1}{16}*\frac{7}{15}$

but in the end i couldn't get the right answer... Any help would be appreciated

Answer 1

answered by the comments and sadly i can't mark them as answers

Perhaps I am mistaken, but it looks like (not A2) and A3 are mutually exclusive (i.e. they can't both happen). If I'm right, either you have a typo or the problem is meaningless or you are supposed to realize this and declare the probability = 0.