A bettor marked ten different numbers on two tickets in the five-number lottery. Four of these numbers were drawn. What is the probability that the bettor had

a) a four-number hit?

b) a two-number hit?
Solution. The 4 numbers can be drawn from the 10 marked numbers in $\binom{10}{4}=210$ ways. This is the total number of cases. Each bettor marked 5 numbers on their ticket; for one of these to have 4 hits, the following scenarios can occur: on one ticket, there are 0 hits, and on the other, there are 4; or vice versa. The number of possibilities is:

$$
\binom{5}{0}\binom{5}{4}+\binom{5}{4}\binom{5}{0}=10
$$

In this case, the probability is:

$$
p=\frac{\text { number of favorable cases }}{\text { total number of cases }}=\frac{10}{210}=\frac{1}{21} \approx 0.0476 \text {. }
$$

Similarly, we can write the probability for the case of two hits as:

$$
\binom{5}{2}\binom{5}{2}=100, \quad \text { and } \quad p=\frac{100}{210}=\frac{10}{21} \approx 0.4762
$$