## Task B-4.5.

A math teacher gave the students ten problems for homework and said that he would select two of them for the next test. How many of the given ten problems does Luka need to solve at minimum to be sure with a probability greater than $\frac{7}{9}$ that at least one of the selected problems will be among those he has solved?
## First Solution.

Let $S$ be a set of ten homework problems. Let $L$ be a subset of $S$, i.e., a set of $x$ problems that Luka has solved. Then $10-x$ is the number of problems that Luka has not solved.

Consider the event $A$ - "the professor has chosen at least one problem from the set $L$."

Then the complementary event $\bar{A}$ is - "the professor has not chosen any problem from the set $L$" or $\bar{A}$ - "the professor has chosen both problems from the set $S \backslash L$."

The probability of event $A$ is

$$
p(A)=1-p(\bar{A})=1-\frac{\binom{10-x}{2}}{\binom{10}{2}}=1-\frac{(10-x)(10-x-1)}{10 \cdot 9}
$$

Since $p(A)>\frac{7}{9}$, it follows that

$$
\begin{gathered}
1-\frac{(10-x)(10-x-1)}{10 \cdot 9}>\frac{7}{9} \\
90-\left(100-10 x-10-10 x+x^{2}+x\right)>70 \\
-x^{2}+19 x-70>0
\end{gathered}
$$

The solution to this inequality is the interval $\langle 5,14\rangle$,

Note: For a student who does not describe the analyzed events in text but correctly solves the problem, do not deduct 1 point.