Answer:
0.9123 or 91.23%
Step-by-step explanation:
Given that none of the selected players are seniors, there are 7 freshmen, 8 sophomore and 6 juniors left (total of 21 players). Since four players are selected, the probability that at most two freshmen are selected is 100% minus the probability that either 4 or 3 freshmen are selected:
[tex]P(X\leq 2)=1-P(X=4)-P(X=3)\\P(X\leq 2)=1-\frac{7}{21}*\frac{6}{20}*\frac{5}{19}*\frac{4}{18}-4*\frac{7}{21}*\frac{6}{20}*\frac{5}{19}*\frac{14}{18}\\P(X\leq 2)=0.9123=91.23\%[/tex]
The probability is 0.9123 or 91.23%
