Question:
$8$ identical balls are randomly distributed into $8$ boxes.
Given first box and second box are not both empty, find the probability that first box is not empty?
$A:=$ B1 is not empty.
$B:=$ B1 and B2 are not both empty.
Answer: $$P(B)= 1-\frac{4^8}{6^8}$$
$$P(A\cap B)=1-\frac{5^8}{6^8}$$
Then, $$P(A\vert B) =\frac{P(A\cap B)}{P(B)}=\frac{6^8-5^8}{6^8-4^8}$$
Answer is this. But, I cannot understand. How is the answer found? -espacially $P(B)=? $ and $P(A\cap B)=?$- Please explain it step by step. Thank you:)
(Assuming you have 8 identical balls and 6 boxes):
Before checking the below answer check the link I gave in the comments above. Also this is the same thing I tried to point in your last question:Two probability questions.
A:= B1 is not empty.
B:= B1 and B2 are not both empty.
Size of sample space = $\binom{13}{5}$
$P(B1\ is\ empty) = \frac{\binom{12}{4}}{\binom{13}{5}} = P(\bar{A})$
$\therefore P(A) = 1 - \frac{\binom{12}{4}}{\binom{13}{5}} = \frac{\binom{12}{5}}{\binom{13}{5}}$
$P(B1\ and\ B2\ are\ both\ empty) = \frac{\binom{11}{3}}{\binom{13}{5}} = P(\bar{B})$
$\therefore P(B) = 1 - \frac{\binom{11}{3}}{\binom{13}{5}} = \frac{\binom{11}{4} + \binom{12}{5}}{\binom{13}{5}}$
$P(A \cap B) = 1 - P(\bar{A} \cup \bar{B})$
Since $\bar{B} \subset \bar{A}$, we can say: $A \subset B$
$\therefore P(A \cap B) = P(A)$
$\therefore P(A|B) = \frac{P(A)}{P(B)} = \frac{\binom{12}{5}}{\binom{11}{4} + \binom{12}{5}}$