Q.4.13 - A First Course in Probability

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 

Step-by-Step Solution

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 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 $)=(1-0.3)\times (1-0.6)=0.7\times 0.4=0.28$

Step 3: 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

$P(X=0)=0.28$

$P(X=500)=0.27$

$P(X=1000)=0.315$

$P(X=1500)=0.09$

$P(X=2000)=0.045$