An urn contains 6 white, 3 red, 2 blue, and 10 green balls. We draw 3 balls one after another. What is the probability that all three drawn balls are white,

a) if the drawn balls are not replaced into the urn?

b) if the drawn ball is replaced into the urn after each draw?

c) What is the probability of drawing 3 white balls if we draw 3 balls at once?
I. solution: Since the probability of drawing each ball is the same, the probability of drawing a ball of a certain color is

$$
v=\frac{\text { number of favorable cases }}{\text { number of possible cases }}
$$

a) The number of possible cases is equal to the number of third-class permutations without repetition from 21 elements $V_{21}^{3}=21 \cdot 20 \cdot 19$. The number of favorable cases is also a third-class permutation, but from 6 elements, so their number is $V_{6}^{3}=6 \cdot 5 \cdot 4$. The sought probability is

$$
V_{a}=\frac{6 \cdot 5 \cdot 4}{21 \cdot 20 \cdot 19}=\frac{2}{133} \approx 0.015
$$

b) The number of possible cases is now $V_{21}^{i, 3}=21^{3}$, and the number of favorable cases is $V_{6}^{i, 3}=6^{3}$, so the sought probability is

$$
V_{b}=\frac{6^{3}}{21^{3}}=\left(\frac{2}{7}\right)^{3}=\frac{8}{343} \approx 0.023
$$

c) The number of possible cases is now $C_{21}^{3}=\binom{21}{3}$ and the number of favorable cases is $C_{6}^{3}=\binom{6}{3}$, so

$$
V_{c}=\frac{\binom{6}{3}}{\binom{21}{3}}=\frac{6 \cdot 5 \cdot 4}{1 \cdot 2 \cdot 3} \cdot \frac{1 \cdot 2 \cdot 3}{21 \cdot 20 \cdot 19}=\frac{6 \cdot 5 \cdot 4}{21 \cdot 20 \cdot 19}=V_{a}
$$

Remark: Since

$$
\frac{V_{k}^{m}}{V_{l}^{m}}=\frac{\frac{V_{k}^{m}}{k!}}{\frac{V_{l}^{m}}{k!}}=\frac{C_{k}^{m}}{C_{l}^{m}}
$$

we can generally state that in the drawing of objects - if the drawn object is not replaced before each draw - it makes no difference from the probability point of view whether we draw a certain number of objects one by one or all at once.

Frank György (Bp., V. Eötvös g. I. o. t.)

II. solution: a) The probability that the first ball is white

$$
V_{A}=\frac{6}{21}=\frac{2}{7}
$$

The probability that the second ball is also white, given that the first one was white

$$
V_{B / A}=\frac{5}{20}=\frac{1}{4}
$$

The probability that the third ball is also white, given that the first two were white

$$
V_{C \mid A B}=\frac{4}{19}
$$

According to the multiplication rule, the sought probability is

$$
V_{A B C}=\frac{2}{7} \cdot \frac{1}{4} \cdot \frac{4}{19}=\frac{2}{133} \approx 0.015
$$

b) In this case, the probability of drawing a white ball is the same in all three draws: $\frac{6}{21}=\frac{2}{7}$. Therefore, the sought probability is

$$
V=\left(\frac{2}{7}\right)^{3}=\frac{8}{343} \approx 0.023
$$

c) The drawing of balls all at once can be thought of as drawing the balls one by one, provided that the drawn ball is not replaced, i.e., the sought probability is the same as in case a).

Lackner Györgyi (Bp., V. 1. sz. textilip. techn. II. o. t.)