Jeff typically makes 75% of his field goals. Steve typically makes 55% of his field goals. Suppose they both have the opportunity to kick 3 field goals. What is the probability Jeff will succeed in making at least 1 field goal? (0.0001) What is the probability Steve will succeed in making at least 1 field goal? (0.0001) What is the probability that both Steve and Jeff will succeed in making at least one field goal? (0.0001) What is the probability that either Steve or Jeff will succeed in making at least one field goal? (0.0001) Given that only one field goal total was scored, what is the probability that Jeff was the one who kicked it? (0.0001) Solution p(jeff) = 0.75 p(steve) = 0.55 p(jess scores atleast 1) = 1 - (0.25)^3 = 0.984375 p(steve scores atleast 1) = 1 - (0.45)^3 = 0.908875 p(both score atleast one goal) = 0.984375 x 0.908875 = 0.89467 p(either jeff or steve score atleast one goal) = 0.984375 + 0.908875 - 0.89467 = 0.99858 p(jeff scores 1 goal and steve none), p1 = 3*0.75*(0.25)^2*(0.45)^3 = 0.01281 p(steve scores 1 goal and jeff none), p2 = 3*0.55*(0.45)^2*(0.25)^3 = 0.00522 thus, p(jeff scores the only goal scored) = p1/(p1+p2) = 0.71.

1. Jeff typically makes 75% of his field goals. Steve typically makes 55% of his field goals.
Suppose they both have the opportunity to kick 3 field goals.
What is the probability Jeff will succeed in making at least 1 field goal? (0.0001)
What is the probability Steve will succeed in making at least 1 field goal? (0.0001)
What is the probability that both Steve and Jeff will succeed in making at least one field goal?
(0.0001)
What is the probability that either Steve or Jeff will succeed in making at least one field goal?
(0.0001)
Given that only one field goal total was scored, what is the probability that Jeff was the one who
kicked it? (0.0001)
Solution
p(jeff) = 0.75 p(steve) = 0.55
p(jess scores atleast 1) = 1 - (0.25)^3 = 0.984375
p(steve scores atleast 1) = 1 - (0.45)^3 = 0.908875
p(both score atleast one goal) = 0.984375 x 0.908875 = 0.89467
p(either jeff or steve score atleast one goal) = 0.984375 + 0.908875 - 0.89467 = 0.99858
p(jeff scores 1 goal and steve none), p1 = 3*0.75*(0.25)^2*(0.45)^3 = 0.01281
p(steve scores 1 goal and jeff none), p2 = 3*0.55*(0.45)^2*(0.25)^3 = 0.00522
thus, p(jeff scores the only goal scored) = p1/(p1+p2) = 0.71