6

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?

N. F. Taussig
  • 76,571
Sebastian Smith
  • 61
  • 2

3 Answers3

13

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:

$$\frac{6}{71}\cdot \frac{1}{14} + \frac{65}{71}\cdot \frac{4}{70}$$

Bram28
  • 100,612
  • 6
  • 70
  • 118
  • This is absolutely correct. But your first sentence is a bit misleading - 'imagine that you first put one white ball randomly in one of the boxes'. This seems to suggest that each box is equally likely. In fact, as your calculations show, you put it in a box with a probability proportional to the number of places in the box. – jwg May 24 '18 at 14:49
  • @jwg Good point! I'll make that more clear, thanks! – Bram28 May 24 '18 at 14:51
11

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.

drhab
  • 151,093
4

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.

Phil H
  • 1,338