What is the probability that white balls will be in the same box? I have trouble computing a probability, I wish someone can help me with this. This is not a homework or anything like that.
So there are $71$ balls, and two of them are white, the rest are black.
The balls are randomly distributed between $14$ boxes. $5$ balls in each box, one box will have $6$ balls. What is the probability that white balls will be in the same box?
 A: I saw this as the set of pairs-in-the-same-box vs all pairs:
$a = (5\mathbb{C}2).13 + 6\mathbb{C}2 = 145\ \text{pairs} \\
b = 71\mathbb{C}2 = 2485\ \text{pairs} \\
\frac{a}{b} = \frac{145}{2485} \approx 5.8\%$
Where $a$ is the collected set of co-located pairs from each of the boxes (summed because they are independent), and $b$ is all the pairs of balls in total.
A: To figure this out, assume there is a 'first' and 'second' white ball, and calculate the chance that the second ball will end up in that same box as the first. So, if we focus on the 'first' ball:
There is a $\frac{6}{71}$ chance it ends up in the box with $6$ balls, and since at that point all of the $14$ boxes have an equal amount of room ($5$ balls) left, there is a $\frac{1}{14}$ chance the second white ball ends up in that box as well. 
However, there is a $\frac{65}{71}$ chance the first ball ends up in one of the other boxes, in which case there is only a $\frac{4}{70}$ chance the second white ball ends up in that same box. 
Hence, the chance of both of them ending up in the same box is:
$$\frac{6}{71}\cdot \frac{1}{14} + \frac{65}{71}\cdot \frac{4}{70}$$
A: If the boxes are numbered $1,\dots,13,14$ and $6$ balls are placed in box $14$ then for $i\in\{1,\dots,13\}$ the probability that both white balls are placed in box $i$ is $\frac5{71}\frac4{70}$.
The probability that both white balls are placed in box $14$ is $\frac6{71}\frac5{70}$.
The described events are mutually exclusive so this results in a probability:$$13\times\frac5{71}\frac4{70}+\frac6{71}\frac5{70}$$that the white balls are placed in the same box.
