## Task 36/73

From an urn containing a total of 53 black and white balls, four balls are drawn at random.

The drawn balls are not returned to the urn. How many balls of each type must be in the urn so that the probability of drawing either four black balls or two black and two white balls is the same?
Assuming there are $a$ black and $b=53-a$ white balls in the urn. Then the probability of drawing four black balls is

$$
W_{4s}=\frac{\binom{a}{4}}{\binom{53}{4}}=\frac{a \cdot(a-1) \cdot(a-2) \cdot(a-3)}{53 \cdot 52 \cdot 51 \cdot 50}
$$

the probability of drawing two black and two white balls is

$$
W_{2s2w}=\binom{4}{2} \frac{\binom{a}{2}\binom{b}{2}}{\binom{53}{2}\binom{51}{2}}=6 \cdot \frac{a \cdot(a-1) \cdot b \cdot(b-1)}{53 \cdot 52 \cdot 51 \cdot 50}
$$

(the factor $\binom{4}{2}=6$ arises from the fact that the combination of black and white balls can occur in $\binom{4}{2}=6$ different ways).

If $a \leq 1$, then $W_{4s}=0$ and $W_{2s2w}=0$, so $W_{4s}=W_{2s2w}$. We now investigate the case $a \geq 2$. From $W_{4s}=W_{2s2w}$ it follows that

$$
a \cdot(a-1) \cdot(a-2) \cdot(a-3)=6 a(a-1) b(b-1) \rightarrow(a-2)(a-3)=6(53-a)(52-a)
$$

This results in the quadratic equation $a^{2}-125 a+3306=0$, which has only the solution $a=38$ in the range $2 \leq a \leq 53$. For $b$, this gives $b=15$.

The probabilities are therefore equal if and only if there are fewer than two or exactly 38 black balls in the urn.

### 2.14 Problems and Solutions 1974