Using the binomial distribution, it is found that 3,845 people can be expected to win in the second round.
[tex]E(X) = np[/tex]
In this problem, two hundred thousand people enter a multistage lottery, hence [tex]n = 200000[/tex].
The odds of winning in the first round are 1 in 50, hence for the first round [tex]p = \frac{1}{1 + 50} = \frac{1}{51}[/tex], and the number of people expected to win is:
[tex]E_1 = 200000\frac{1}{51} = 3922[/tex]
Hence, in the second round, the number of people contending to win will be:
[tex]n = 200000 - 3922 = 196078[/tex]
The odds of winning the second round are 2 in 100, hence:
[tex]p = \frac{2}{100 + 2} = \frac{2}{102} = \frac{1}{51}[/tex]
Then, the number of people expected to win is:
[tex]E_2 = 196078\frac{1}{51} = 3845[/tex]
3,845 people can be expected to win in the second round.
You can learn more about the binomial distribution at https://brainly.com/question/14424710
