# Answer on Statistics and Probability Question for mike

Question #26450

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

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

Expert's answer

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%

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%

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