Greg has a .78 chance of making par on each hole of golf that he plays. Today he plans to play just three holes. Find the probability of each event. Round answers to three significant figures.

Question

a) He makes par on all three holes.

b) He makes par of exactly two of the three holes.

c) He makes par on at least one of the three holes.

d) He makes par on the first and third holes but not the second.

Answer 1

This is a great problem to use a tree diagram:

This answer format doesn't allow me to paste in the tree diagram I made because it takes up too much memory but the results are these:

Where "P" is Par and "N" is Not Par these are the possible results:

PPP

PPN

PNP

PNN

NPP

NPN

NNP

NNN

The probabilities that each will occur is the multiplication of the individual probabilities:

PPP: (0.78)(0.78)(0.78) = 0.475

PPN: (0.78)(0.78)(0.22) = 0.134

PNP: (0.78)(0.22)(0.78) = 0.134

PNN: (0.78)(0.22)(0.22) = 0.038

NPP: (0.22)(0.78)(0.78) = 0.134

NPN: (0.22)(0.78)(0.22) = 0.038

NNP: (0.22)(0.22)(0.78) = 0.038

NNN: (0.22)(0.22)(0.22) = 0.011

1) The probability he makes par on all three holes: 47%

2) The probability he makes par of exactly two of the three holes: 0.134•3 = 40%

3) The probability he makes par on at least one of the three holes: 1 - .011 = 99%

4) The probability he makes par on the first and third holes but not the second: 13%