Question #278223

1. a) Two IPod salesmen A and B must each make two calls per day, one in the morning and in the afternoon. A has a probability 0.4 of selling a laptop on call, while B has a probability 0.1 of a sale. A works independently of B, and for each salesman, morning and afternoon are independent of each other.

Find the probability that, in one day:


i) A sells two iPods [2 Marks]


ii) A sells just one iPod [2 Marks]


iii) B makes at least one sale [3 Marks]


iv) Between them A and B make exactly one sale. [4Marks]




1
Expert's answer
2021-12-13T11:10:38-0500

i)


P(AA)=0.4(0.4)=0.16P(AA)=0.4(0.4)=0.16


ii)


P(A only 1)=P(AAC)+P(ACA)P(A\ only\ 1)=P(AA^C)+P(A^CA)

=0.4(10.4)+(10.4)(0.4)=0.48=0.4(1-0.4)+(1-0.4)(0.4)=0.48

iii)


P(B at least 1)=1P(BCBC)P(B\ at\ least\ 1)=1-P(B^CB^C)

=1(10.1)(10.1)=0.19=1-(1-0.1)(1-0.1)=0.19

iv)


P(only 1)=P(A only 1)P(BCBC)P(only\ 1)=P(A\ only\ 1)P(B^CB^C)

+P(B only 1)P(ACAC)+P(B\ only\ 1)P(A^CA^C)

=(P(AAC)+P(ACA))P(BCBC)=(P(AA^C)+P(A^CA))P(B^CB^C)

+(P(BBC)+P(BCB))P(ACAC)+(P(BB^C)+P(B^CB))P(A^CA^C)

=(0.4(10.4)+(10.4)(0.4))(10.1)(10.1)=(0.4(1-0.4)+(1-0.4)(0.4))(1-0.1)(1-0.1)

+(0.1(10.1)+(10.1)(0.1))(10.4)(10.4)+(0.1(1-0.1)+(1-0.1)(0.1))(1-0.4)(1-0.4)

=0.4536=0.4536

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