Lenka chose $4$ distinct integers from the interval $[1, 19]$, and is solving for the probability that their sum will be an even number.
Lenka reasoned that there are only $3$ ways to select the integers to achieve an even sum: either all $4$ numbers will be even, the second possibility is $2$ even and $2$ odd numbers, and the last possibility is $4$ odd numbers.
(1) First, Lenka calculated the number of draws in which all $4$ numbers are even: "There are $9$ even numbers in the given interval, and the number of draws of $4$ even numbers from $9$ is ${9 \choose 4}=126$.“
(2) Next, she calculated the number of draws in which exactly two numbers are even and exactly two numbers are odd: "There are $9$ even numbers and $10$ odd numbers in the given interval. The number of such draws is ${9 \choose 2}+{10 \choose 2}=81$.“
(3) She calculated the number of draws in which all $4$ numbers are odd: "There are $10$ odd numbers in the given interval. Therefore, the number of such selections is ${10 \choose 4}=210$.“
(4) She determined the number of selections of $4$ integers that satisfy the condition that their sum is even: $126 + 81 + 210 = 417$.
(5) She calculated the total number of possible selections of $4$ numbers from $19$ as ${19 \choose 4}=3876$.
(6) She determined the probability that when randomly selecting $4$ integers from the interval $[1, 19]$, their sum will be an even number, as $\frac{417}{3876}\approx 0.1076$.
Lenka solved the problem correctly.
Lenka made a mistake in step (1). The number of draws of $4$ even numbers is $4\cdot9 = 36$. The correct probability is $\frac{36 + 81 + 210}{3876} \approx 0.0844$.
Lenka made a mistake in step (2). The number of draws with two even and two odd numbers is ${9\choose 2}\cdot{10 \choose 2} = 1620$. The sought probability is $\frac{126 + 1 620 + 210}{3876}\approx 0.5046$.
Lenka made a mistake in calculating the probability in step (6). The correct probability is $1 – \frac{417}{3876}\approx0.8924$.