6. Suppose that there is a gambling game, where there is a 60% probability that

Question

6. Suppose that there is a gambling game, where there is a 60% probability that you will lose $1, and a 40% proba bility that you will gain S1. Suppose that you play the game 100 times, and each time you play is independent of any other time (a) Find the probability the total amount you go home with is at least 0 (you have broken even, or went home with some positive gain) (b) Find the probability that the total amount you go home with is larger that $5 (you have a net gain of $5) (c) Find the probability that the total amount you go home with is less that -S5 (you have a net loss of $5) (d) Find the 99th percentile for the total amount you go home with

Explanation / Answer

here expected gain per game E(X)=(-1)*0.6+1*0.4=-0.2

E(X2) =(-1)2*0.6+12*0.4=1

std deviation=(E(X2)-(E(X))2)1/2 =(1-(-0.2)2)1/2 =0.9798

therefore for 100 games expected gain =100*(-0.2)=-20

std deviation =0.9798*(100)1/2 =9.798

for normal approximation z score =(X-mean)/std deviaiton

a)

P(X>0) =1-P(X<0) =1-P(Z<(0-(-20)/9.798)=1-P(Z<2.04)=1-0.9793 =0.0207

b)

P(X>5) =1-P(X<5) =1-P(Z<(5-(-20)/9.798)=1-P(Z<2.55)=1-0.9946 =0.0054

c)

P(X<-5) =P(Z<(-5-(-20)/9.798)) =P(Z<1.53)=0.9370

d)

for 99th percentile critical z =2.33

hence corresponding value =mean +z*std deviaiton =-20+2.33*9.798=2.829

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