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In a population of 200 students who have just completed a first course in calculus, 50 have earned A's, 80 B's and remaining earned F's. A sample of size 25 is taken at random and without replacement from this population. What is the probability that 10 students have A's, 12 students have B's and 3 students have F's?

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  • $\begingroup$ I did $P(\text{10 A's, 12 B's, and 3 F's}) = P(X=A)^10 P(X=B)^12 P(X=F)^3$ but considering that a sample size of 25 is taken, we must consider it $\endgroup$ – geniwebb May 14 '17 at 14:19
  • $\begingroup$ Your approach would work if we picked the students with replacement, i.e. if nothing stops the same student from being picked more than once. However, that's not the case here. $\endgroup$ – Arthur May 14 '17 at 14:42
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There are $\binom{200}{25}$ many possible selections of 25 from the 200 students.

To get the desired distribution we need $10$ out of the $50$ $A$'s, which can happen in $\binom{50}{10}$ ways. Next we independently pick $12$ out of the $80$ $B$'s which can happen in $\binom{80}{12}$ ways and finally $3$ out of the remaining $70$ again (in ...ways). Now multiply and divide.

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