The state runs a lottery once every week in which six number

Step 1
Hyper geometric distribution:
A hyper geometric distribution is a discrete probability distribution that determines the probability of getting k successes in n draws (without replacement) from a finite population of size N that contains exactly K success states.
Denote the total number of successes as k,
Denote the total number of objects that are drawn without replacement as n,
Denote the population size as N,
Denote the total number of success states in the population as K.
The probability distribution of k is a hyper geometric distribution with parameters (N, K, n) and the probability mass function (pmf) of k is given as:
$$P(k)=\frac{(\begin{array}{c}K\\ k\end{array})\times (\begin{array}{c}N-K\\ n-k\end{array})}{(\begin{array}{c}N\\ n\end{array})}$$
Step 2
Parameters of hyper geometric distribution for the given situation:
In a state’s lottery, 6 numbers are randomly selected from a total of 17 numbers without replacement.
Here, the population size is ${\displaystyle N=17}$,
The number of draws is ${\displaystyle n=6}$.
Furthermore, it is given that, a player chooses 6 numbers before the state’s sample is selected and the player wins, if all the 6 numbers match with the state’s sample.
Thus, the number of success states in the population is ${\displaystyle K=6}$.
The probability distribution of k successes is a hyper geometric distribution with parameters (${\displaystyle N=17,K=6,n=6}$) and the probability mass function (pmf) of k is given as:
$$P(k)=\frac{(\begin{array}{c}6\\ k\end{array})\times (\begin{array}{c}17-6\\ 6-k\end{array})}{(\begin{array}{c}17\\ 6\end{array})}$$
Step 3
Probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample:
It is given that, 3 of the 6 chosen numbers have to appear in the state’s lottery.
That is, the number of successes is ${\displaystyle k=3}$.
The probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample is obtained as 0.267 from the calculation given below:
$$P(k=3)=\frac{(\begin{array}{c}6\\ 3\end{array})\times (\begin{array}{c}17-6\\ 6-3\end{array})}{(\begin{array}{c}17\\ 6\end{array})}$$
${\displaystyle =\frac{20\times 165}{12,376}}$
${\displaystyle =\frac{3,300}{12,376}}$
${\displaystyle =0.2666}$
${\displaystyle =0.267}$ (Rounded to 3 decimal places)
Step 4
Answer:
The probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample is 0.267.