A probability problem by Aira Thalca

Probability Level 4

6 different balls are to be placed in three different boxes A, B, C. Each ball is placed uniformly at random in one of these boxes. The probability that box A will contain exactly 3 balls can be written as $\frac{a}{b}$ , where $a$ and $b$ are coprime positive integers. Find $a+b$ .

The answer is 889.

1 solution

James Pohadi
Dec 31, 2016

Each ball can be placed randomly in $3$ different boxes. So, the total possible cases $P(s)$ is $3^6$ .

Box $A$ contain exactly $3$ balls from $6$ balls, the remaining $3$ balls can be placed randomly in the other $2$ boxes. The total possible cases $P(n)$ is $6C3 \times 2^3$ .

$P= \dfrac{P(n)}{P(s)}=\dfrac{6C3 \times 2^3}{3^6}=\dfrac{160}{729}$

$a=160$ , $b=729$ . Then $a+b=160+729=\boxed{889}$

Jup did it the same way. It is just binomially distributed with p=1/3 and n=6.

Peter van der Linden - 4 years, 5 months ago

A probability problem by Aira Thalca

The answer is 889.

1 solution

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