Uma pequena faculdade tem 800 alunos, $10 \%$ dos quais são canhotos. Suponha que eles tomem uma amostra de AAS 4 alunos. Seja $X=$ o número de alunos canhotos na amostra.

Qual é a probabilidade de que exatamente 2 dos 4 alunos sejam canhotos?

Você pode arredondar sua resposta para a segunda casa decimal.
\[
P(X=2)=
\]

Step 1 ：The problem is asking for the probability that exactly 2 out of 4 students are left-handed. This is a binomial probability problem, where the number of trials is 4 (the number of students sampled), the number of successes we want is 2 (the number of left-handed students), and the probability of success on each trial is 0.10 (the proportion of left-handed students in the population).

Step 2 ：The formula for binomial probability is: \[P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\] where: \[P(X=k)\] is the probability we are trying to find, \[C(n, k)\] is the number of combinations of n items taken k at a time, \[p\] is the probability of success on each trial, \[n\] is the number of trials, and \[k\] is the number of successes we want.

Step 3 ：In this case, n=4, k=2, and p=0.10. We can plug these values into the formula to find the probability.

Step 4 ：n = 4, k = 2, p = 0.1, combinations = 6, probability = 0.05

Step 5 ：Final Answer: The probability that exactly 2 out of 4 students are left-handed is \(\boxed{0.05}\).