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

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


A
Asked by 2 years ago
45 points

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. Determine the probability mass function of X, the total dollar value of all sales.

A salesman
Adrian

1 Answer

Answered by 2 years ago
8.7k points

There are a few scenarios which we should consider. The possibility that the salesman sells no encyclopedias, the \$500 encyclopedia, or the \$1000 encyclopedia. So there are $3^2=9$ different ways to make these sells. The possible scenarios for selling to the first and second customer looks like,

${(0,0),(0,500),(0,1000),(500,0),(500,500),(500,1000),(1000,0),(1000,500),(1000,1000)}$

where our random variable X is the sum of the sales to the first and second customer

let

F = first appointment sale
S = second appointment sale

furthermore,

$F_1$= sale of \$500 on first appt
$F_2$= sale of \$1000 on first appt
$S_1$= sale of \$500 on second appt
$S_2$= sale of \$1000 on second appt

finally,

$P(F)=0.3, P(F^c)=1-0.3=0.7$
$P(S)=0.6, P(S^c)=1-0.6=0.4$
$P(F_1|F)=P(F_2|F)=P(S_1|S)=P(S_2|S)=0.5$ $P(F_1)=P(F_2)=P(F_1|F)P(F)+P(F_1|F^c)P(F^c)=(0.5)(0.3)+(0)(0.7)=0.15$
$P(S_1)=P(S_2)=P(S_1|S)P(S)+P(S_1|S^c)P(S^c)=(0.5)(0.6)+(0)(0.4)=0.3$

consider a few,

$$P(X=0)=P(F^c \cap S^c)=P(F^c)P(S^c)=(1-0.3)(1-0.6)=0.28$$

$$P(X=500)=P((500,0))+P((0,500))=P(F_1)P(S^c)+P(F^c)P(S_1)$$

$$=(0.15)(0.4)+(0.7)(0.3)=0.27$$

continue after this manner to find P(X=1000), P(X=1500), P(X=2000) making sure your probabilities sum to 1.

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Asked: 2 years ago

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