Answer to Question #26450 in Statistics and Probability for mike

Question

greg has a 78% chance of making par on each hole he decides to play in golf. Today he plans to play just three holes. Find the probability that..
A greg makes par on all three holes
Bgreg makes does not make par on any hole
c greg makes par on at least one hole
D greg makes par on the first and third hole, but not the second hole

Answer 1

Greg has a 78% chance of making par on each hole he decides to play in golf. Today he plans to play just three holes.& Find the probability that:

A Greg makes par on all three holes
Probability of making par on one hole:
P(1) = 78% = 0.78
Therefore, Probability of making par on three holes:
P(3) = P(1)^3 = 0.78^3 = 0.475 = 47.5%
Answer: 47.5%

B Greg makes does not make par on any hole
Probability of making par on one hole:
P(1) = 78% = 0.78
Probability of not making par on one hole:
P(-1) = 1 - P(1) = 1 - 0.78 = 0.22
Probability of not making par on three holes:
P(-3) = P(-1)^3 = 0.22^3 = 0.011 = 1.1%
Answer: 1.1%

C Greg makes par on at least one hole
Probability of not making par on three holes:
P(-3) = 1.1%
In some other case he makes par on at least one hole, therefore probability makes par on at least one hole:
P(!0) = 1 - P(-3) = 98.9 %

D Greg makes par on the first and third hole, but not the second hole
Probability of making par on first hole:
P1 = 0.78
Probability of not making par on second hole:
P2 = 1 - 0.78 = 0.22
Probability of making par on third hole:
P3 = 0.78
Therefore, total probability of this events:
P = P1*P2*P3 = 0.78^2*0.22 = 0.134 = 13.4%
Answer: P=13.4%