The gift shop at Emory Midtown Hospital has kept track of the purchases of newsp

Question

The gift shop at Emory Midtown Hospital has kept track of the purchases of newspapers and magazines by customers over the last week. Let X = the number of magazines purchased and let Y = the number of newspapers published. Each of the customers purchased 0, 1 or 2 newspapers and 0, 1, 2, or 3 magazines. There were 100 total customers last week, and the number of customers who purchased each possible combination of newspapers and magazines is given in the following table: a) What is the probability that a random customer did not purchase any newspapers or magazines? b) What do we call the approach used to determine the probability asked for in part a? c) Create a table that shows the probability that X takes on each of its possible values. d) Create a table that shows the probability that Y takes on each of its possible values. e) What is the probability that a customer purchased more newspapers than magazines? f) What is the probability that a customer did not purchase any magazines given that he or she purchased 1 or 2 newspapers? g) What is the expected number of newspapers purchased by each customer and what is the expected number of magazines purchased by each customer? h) Say that newspapers cost $.60 each and magazines cost $2 each. What is the expected amount spent on newspapers and magazines by each customer?

Explanation / Answer

here total number of customers=100

X=number of magazines purchased

Y=number of newspaper purchased

a) we want to find the probability that a random customer did not buy any newspaper or magazine

so required probability is P[X=0 or Y=0]=P[X=0]+P[Y=0]-P[X=0 and Y=0]

now P[X=0]=(28+12+5)/100=45/100

P[Y=0]=(28+15+5+2)/100=50/100

P[X=0 and Y=0]=28/100

so required probability is P[X=0 or Y=0]=45/100+50/100-28/100=67/100=0.67 [answer]

b) this method is called using Poincare's theorem.

c) X can take the values 0,1,2,3

P[X=0]=P[X=0,Y=0]+P[X=0,Y=1]+P[X=0,Y=2]=28/100+12/100+5/100=45/100=0.45

P[X=1]=P[X=1,Y=0]+P[X=1,Y=1]+P[X=1,Y=2]=15/100+10/100+5/100=30/100=0.30

P[X=2]=P[X=2,Y=0]+P[X=2,Y=1]+P[X=2,Y=2]=5/100+9/100+3/100=17/100=0.17

P[X=3]=P[X=0,Y=0]+P[X=0,Y=1]+P[X=0,Y=2]=2/100+4/100+2/100=8/100=0.08

hence the table is

0

d) Y takes the values 0,1,2

so P[Y=0]=P[Y=0,X=0]+P[Y=0,X=1]+P[Y=0,X=2]+P[Y=0,X=3]=28/100+15/100+5/100+2/100=50/100=0.5

P[Y=1]=P[Y=1,X=0]+P[Y=1,X=1]+P[Y=1,X=2]+P[Y=1,X=3]=12/100+10/100+9/100+4/100=35/100=0.35

P[Y=2]=P[Y=2,X=0]+P[Y=2,X=1]+P[Y=2,X=2]+P[Y=2,X=3]=5/100+5/100+3/100+2/100=15/100=0.15

hence the table is

e) here the required probability is

P[Y>X]=P[Y=2,X=0]+P[Y=2,X=1]+P[Y=1,X=0]=5/100+5/100+12/100=22/100=0.22 [answer]

f) reqquired probability is

P[ the customer did not buy any magazine given that he has bought 1 or 2 newspapers]

=P[X=0 |Y=1 or 2]=P[X=0 and Y=1 or 2]/P[Y=1 or 2]={P[X=0 and Y=1]+P[X=0 and Y=2]}/{P[Y=1]+P[Y=2]}=

(12/100+5/100)/(0.35+0.15)=0.34 [answer]

g) expected number of newspaper purchased by each customer is

E[Y]=0*0.5+1*0.35+2*0.15=0.65 [answer]

expected number of magazines purchased by each customer is

E[X]=0*0.45+1*0.30+2*0.17+3*0.08=0.88 [answer]

h) cost of newspaper is $0.60 and cost of magazine is $2

hence expected amount spent on nespaper and magazines by each customer is 2*0.88+0.60*0.65=$2.15 [answer]

X P[X=x]

0

0.45 1 0.30 2 0.17 3 0.08

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