## Task A-4.4.

In a box, there is one red and five white balls marked with the numbers $1,2,3,4$, 5. Without looking, Domagoj draws one ball at a time until he draws the red ball, after which he stops drawing. The drawn balls are not returned to the box. What is the probability that the sum of the numbers on the drawn white balls is at least 10?
## First Solution.

Let $A_{k}$ be the event: exactly $k$ balls are drawn ($k-1$ white and one red). Let $B$ be the event: the sum of the numbers on the drawn white balls is at least 10. Then, according to Bayes' formula,

$$
\begin{aligned}
P(B)= & P\left(B \mid A_{1}\right) P\left(A_{1}\right)+P\left(B \mid A_{2}\right) P\left(A_{2}\right)+P\left(B \mid A_{3}\right) P\left(A_{3}\right) \\
& +P\left(B \mid A_{4}\right) P\left(A_{4}\right)+P\left(B \mid A_{5}\right) P\left(A_{5}\right)+P\left(B \mid A_{6}\right) P\left(A_{6}\right)
\end{aligned}
$$

If the red ball is drawn in the first three draws, the sum can be at most $4+5=9$, and if it is drawn in the 5th or 6th draw, the sum will be at least $1+2+3+4=10$. Therefore,

$$
P\left(B \mid A_{1}\right)=P\left(B \mid A_{2}\right)=P\left(B \mid A_{3}\right)=0 \quad \text{and} \quad P\left(B \mid A_{5}\right)=P\left(B \mid A_{6}\right)=1
$$

If the red ball is drawn in the fourth draw, then the sum is at least 10 only if the drawn balls have the numbers $1,4,5$ or $2,3,5$ or $2,4,5$ or $3,4,5$. Each of these four triplets can appear in any of the 3! orders. Therefore,

$$
P\left(B \mid A_{4}\right)=\frac{4 \cdot 3!}{5 \cdot 4 \cdot 3}=\frac{2}{5}
$$

Let's calculate the probability that exactly $k$ balls are drawn, i.e., $P\left(A_{k}\right)$.

For $k=1$, this probability is $\frac{1}{6}$ because out of 6 possibilities, only one is for this event, which is drawing the red ball immediately.

For $k>1$, in the first draw, we can draw any of the 5 white balls out of a total of 6 balls, in the next draw, we can draw any of the remaining 4 white balls out of a total of 5 balls, and so on until in the $(k-1)$-th draw, we draw one of the remaining $5-k+1$ white balls out of a total of $5-k+2$ balls. Finally, we must draw the red ball from the remaining $5-k+1$ balls. Therefore, for all $k=1,2,3,4,5,6$,

$$
P\left(A_{k}\right)=\frac{5}{6} \cdot \frac{4}{5} \cdots \frac{5-k+1}{5-k+2} \cdot \frac{1}{5-k+1}=\frac{1}{6}
$$

The final solution is

$$
P(B)=\frac{2}{5} \cdot \frac{1}{6}+1 \cdot \frac{1}{6}+1 \cdot \frac{1}{6}=\frac{2}{5}
$$