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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:)

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  • Did you get the above solution from behind the book? – Swapniel Mar 01 '14 at 16:12
  • No. From my notebook. @Swapniel – 1190 Mar 01 '14 at 16:55
  • First I would ask how did you get $6^8$ in the denominator? In the question did you mean to distribute 8 balls among 6 boxes? Please check the question. You should have had $8^8$ in the denominator. Second point is, that still the size of you sample space(the denominator is not correct). Take a small case. How many ways can you put 2 identical balls in 2 boxes? Is it 4 ways or 3? Look at the link for reference: http://math.stackexchange.com/questions/47345/number-of-ways-of-distributing-n-identical-objects-among-r-groups – Swapniel Mar 01 '14 at 21:09

1 Answers1

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(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: $A \subset B$
$\therefore P(A \cap B) = P(A)$

$\therefore P(A|B) = \frac{P(A)}{P(B)} = \frac{\binom{12}{5}}{\binom{11}{4} + \binom{12}{5}}$

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