Suppose the probability that a golfer makes a holeinone on t

Suppose the probability that a golfer makes a hole-in-one on the sixth hole of a mini golf course is 1/300. Suppose 36 students play the sixth hole during a round of mini golf. Let x be the number of the 36 who make a hole in one. In how many different ways could exactly 3 students make a hole-in-one on the sixth hole.

The answer is 7,140. But how do you get that answer?

Solution

X be the number of students out of 36 students who make a hole in one.

probability that a golfer makes a hole-in-one on the sixth hole of a mini golf course is 1/300

hence X~Bin(36,1/300)

so the pmf of X is P[X=x]=³⁶C_x(1/300)^x(1-1/300)^36-x       x=0,1,2,....,36

now 3 students out of 36 students can amek a hole-im-one on the sixth whole in

³⁶C₃ ways. that is 3 students are being chosen randomly from 36 students.

now ³⁶C₃=36!/[3!*(36-3)!]=36!/[3!*33!]=36*35*34*33!/[3!*33!]=36*35*34/3!=36*35*34/6=35**34*6=7140 [answer]

and the probability that 3 students make a hole-in-one on the sixth hole is

P[X=3]=³⁶C₃(1/300)³(1-1/300)^36-3= 0.0002368