Question

Can you please explain step by step solution with algebra included, im having a hard time understanding it! Part A & B

7.1.40 In a youth basketball league, a (a) Lat L denote the event that the layup is good and T denote the event that player is fouled in the act of shooting a the the sod t l a layup. There is a probability 0.2 that tree is the layup is good, scoring 2 points. If the layup is good, the player is also awarded 1 free throw, giving the player a chance at a three-point play. If the layup is missed then (because of the foul) the player is still awarded one point automatically and is also awarded one free throw, enabling a este the compkeneatlary evate. The events. The chance to score two points in total. The Reading from the troe, we see that player makes a free throw with probability PIX-1] = (0.8)(0.45-0.4 PIX = 21-(0.2)(0.5) + (08)(0.5) = 0.5, PIX-3-(0.2)(0.5)-0.1. (a) What is the PMF of X, the number of The complete PMF points scored by the player? (b) Find the conditional PMF Pxz() of 0.5 2. X given event T that the free throw is good 10.13 0 otherwise. (b) There are only two outcomes LT (where X = 3) and LT (where X = 2) thn for which eveat T occurs. Since PT (0.2)0.3)+ (0805)a.5, the definition of conditional probability says (08 1-2

Answer 1

(a)This is a question based on probability distribution functions.It was asked to find the Probability Mass Function(PMF) for the event called points scored by a player.Let's X be the points scored by a player.To find it let's see the possible outcomes. Case 1: If the player's layup is good(probability =0.2), He will get minimum 2 points as given question along with a free throw. If his free throw is success (probability = 0.5), His score will be three points.(2+1=3) Otherwise (if free throw is a failure,probability=1-0.5=0.5), His score will be two points only. Case 2: If the player's layup is not good(foul case,probability =1-0.2=0.8),He will get minimum 1 point along with a free throw. If his free throw is success (probability =0.5), His score will be two points.(1+1=2) Otherwise(If free throw is a failure, probability =1-0.5=0.5), His score will be one point only.

Therefore,by observing all the four cases,the possible outcomes are X=1,2,3 only.So the probabilities for these outcomes are P(X=1) = 0.8*0.5 = 0.4 ; P(X=2) = (0.2*0.5)+(0.8*0.5) = 0.5 ; P(X=3) = 0.2*0.5 = 0.1. Hence the P.M.F. is defined as, P_x(x) ={0.4, for x=1 ; 0.5, for x=2 ; 0.1, for x=3 ; 0, otherwise.

(b) It's asked to find conditional probability given free throw is good.In this case there will be two outcomes only.They are 1. layup is good : score is 3 points. (probability = 0.2) 2. layup is not good(foul) : score is 2 points.(probability= 1-0.2=0.8) The conditional P.M.F. can be written as P_{X/Throw is successful}(x) ={0.8, for x=2 ; 0.2, for x=3 ; 0, otherwise.