Jack and Bill are randomly picking students to play kick-ball on the playground, one at a time (without return), from a group of ten students. Only five of the students among the twenty students are athletic. Jack picks first, then Bill, and so on.... What is the probability that Jack will get all of the athletes? (Answer in lowest terms). Pick
A) 1/3	B) 7/273	C) 1/2	D) 1/75	E) 1/252
There are 10 students of which 5/10 are athletes, so the probability of getting an athlete is P(athlete) = 1/2 on the first selected student.   Because selection is (without return), the P(athlete) has changed for following selections. Now, there are 9 students and number athletes is dependent upon whether first student was an athlete, or not... and so on.... Therefore, conditional probability! The probability that event B occurs, given that event A has already occurred is:   Formula = P(A\B) = P(A and B)/P(B).   Since, Jack picks first P(athlete) = 5/10, then Bill picks, and P(not athlete) = 5/9, next P(athlete) = 4/8, etc....   P(A\B) = (5/10)(5/9)(4/8)(4/7)(3/6)(3/5)(2/4)(2/3)(1/2) = 14400/3628800 = 1/252   Answer: 1/252
<result>E</result>
Jack and Bill are randomly picking students to play kick-ball on the playground, one at a time (without return), from a group of ten students. Only five of the students among the twenty students are athletic. Jack picks first, then Bill, and so on.... What is the probability that Jack will get all of the athletes? (Answer in lowest terms).