An urn contains 15 red, 9 white, and 4 green balls. If we draw 3 balls one after another without replacement, what is the probability that

a) the first is red, the second is white, and the third is green?

b) the 3 drawn balls are red, white, and green, regardless of the order?
a) The probability for Anna to first draw a red is $v_{A}=\frac{15}{28}$, to secondly draw a white is $v_{B / A}=\frac{9}{27}$, and to thirdly draw a green is $v_{C / A B}=\frac{4}{26}$. According to the multiplication rule, the sought probability is

$$
v_{a}=v_{A B C}=\frac{15 \cdot 9 \cdot 4}{28 \cdot 27 \cdot 26}=\frac{5}{182} \approx 0.027
$$

b) The probability of drawing any permutation of the red, white, and green elements is the same as the previously calculated $v_{a}$, because the factors in the numerator and denominator remain the same, only permuting in the numerator. Therefore, the sought probability is

$$
v_{b}=3!v_{a}=6 \cdot \frac{5}{182}=\frac{15}{91} \approx 0.165
$$

Török István (Csorna, Latinka Sándor g. I. o. t.)