Question

A salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to a sale with probability $.3,$and his second will lead independently to a sale with probability $.6$. Any sale made is equally likely to be either for the deluxe model, which costs $\backslash (1000$, or the standard model, which costs $\backslash )500.$ Determine the probability mass function of $X$, the total dollar value of all sales

Short answer

$P(X=0)=0.28$

$P(X=500)=0.27$

$P(X=1000)=0.315$

$P(X=1500)=0.09$

$P(X=2000)=0.045$

Step-by-step solution

Step 1:Given information

A salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to a sale with probability .$3$, and his second will lead independently to a sale with probability $.6$. Any sale made is equally likely to be either for the deluxe model, which costs $\$1000$, or the standard model, which costs $500

Step 2:Explanation

Given,

$P($sale $\mid$first $)=0.3$

$P($sale$\mid$second $)=0.6$

$P($deluxe $\mid$sale $)=0.5$

$P($standard $\mid$sale $)=0.5$

$X$is the total dollar value of all sales, which is either $0,500,1000,1500$, or $2000$(because the two sales are worth either $0,500$or $1000$).

$P(X=0)=P($no sale $\mid$ first $)\times P($ no sale $\mid$ second$)=(1-0.3)\times (1-0.6)=0.7\times 0.4=0.28$

Explanation

$P(X=500)=P($no sale $\mid$first $)\times P($sale $\mid$second $)\times P($standard $\mid$sale $)$

$+P($sale $\mid$first $)\times P($no sale $\mid$second$)\times P($standard $\mid$sale $)$

$=(1-0.3)\times 0.6\times 0.5+0.3\times (1-0.6)\times 0.5=0.27$

Step 4:Explanation

$P(X=1000)=P($sale$\mid$first $)\times P($sale $\mid$second $)\times P($standard $\mid$sale $)\times P($standard $\mid$sale $)$

$+P($sale $\mid$first $)\times P($no sale $\mid$second $)\times P($deluxe$\mid$sale $)$

$+P($no sale $\mid$first $)\times P($sale$\mid$second $)\times P($deluxe$\mid$sale$)$

$=0.3\times 0.6\times 0.5\times 0.5+0.3\times (1-0.6)\times 0.5+(1-0.3)\times 0.6\times 0.5=0.315$

Step 5:Explanation

$P(X=1500)=P($sale $\mid$first $)\times P($sale $\mid$second $)\times P($standard$\mid$sale $)\times P($deluxe$\mid$sale$)$

$+P($sale $\mid$first $)\times P($sale $\mid$second $)\times P($deluxe $\mid$sale $)\times P($standard $\mid$sale $)$

$=0.3\times 0.6\times 0.5\times 0.5+0.3\times 0.6\times 0.5\times 0.5=0.09$

Step 6:Explanation

$P(X=2000)=P($sale $\mid$first $)\times P($sale $\mid$second $)\times P($deluxe$\mid$sale $)\times P($deluxe $\mid$sale$)$

$=0.3\times 0.6\times 0.5\times 0.5=0.045$

Step 7:Final answer

$P(X=0)=0.28$

$P(X=500)=0.27$

$P(X=1000)=0.315$

$P(X=1500)=0.09$

$P(X=2000)=0.045$