I. In a basketball game, a team can score a goal (event G) in three ways: F-free
ID: 3306721 • Letter: I
Question
I. In a basketball game, a team can score a goal (event G) in three ways: F-free throw for 1 point), N = { normal two point shot), and T-(three-point shot). For a certain team, we have the following shooting percentages: P(GF) = 0.7, P{ GIN} = 0.4, and P(GIT)-02. On average, the number of shots taken during a typical game, is 20%F, 60%N, and 2096T (a) Estimate the number of points made by this team on three-point shots in a typical game. (b) If the ball is shot, what is the probability that a goal is made? (c) If a goal is made, what is the probability it was made on a normal shot? (d) If a goal is missed, what is the probability it was either a free throw or a three-point shot?Explanation / Answer
P(G|F) = 0.7 P(G|N) = 0.4 P(G|T) = 0.2
P(F) = 0.2 P(N) = 0.6 P(T) = 0.2
(a) Number of points made on three point shots = 3 * P(T) * P(G|T) = 3 * 0.2 * 0.2 = 3 * 0.04 = 0.12
(b) If the ball is shot, the probability that it is a two point shot = 0.6/0.8
Probability that a goal is made on the two point shot = 0.4
If the ball is shot, the probability that it is a two point shot = 0.2/0.8
Probability that a goal is made on the two point shot = 0.2
=> Probability that a goal is made if a ball is shot = 0.6/0.8 * 0.4 + 0.2/0.8 * 0.2
= 0.35
(c) P(G F) = P(G|F) P(F)
= 0.7 * 0.2
= 0.14
P(G N) = P(G|N) P(N)
= 0.4 * 0.6
= 0.24
P(G T) = P(G|T) P(T)
= 0.2 * 0.2
= 0.04
P(G) = 0.14 + 0.24 + 0.04 = 0.42
P(N|G) = P(G N) / P(G)
= 0.24 / 0.42
= 0.57
(d) P(Gc) = 1 - 0.42 = 0.58
P(Gc F) = P(F) - P(G F)
= 0.2 - 0.14
= 0.06
=> P(F|Gc) = P(Gc F) * P(Gc) = 0.06 * 0.58 = 0.0348
P(Gc T) = P(T) - P(G T)
= 0.2 - 0.04
= 0.16
=> P(T|Gc) = P(Gc T) * P(Gc) = 0.16 * 0.58 = 0.0928
=> Probability that the goal made was on either a free throw or a three point shot
= 0.0348 + 0.0928
= 0.1276