Question 1117767: Please help me solve thhis question: A salesperson has a policy of calling upon a customer a second time if she does not make the sale on the first call, and trying a third time, and last time if she fails on the first two calls. She has a probability of a sale on the first call of 0.3. If she fails on the first call, the probability of a sale on the second call is 0.2. If the first two calls are unsuccessful, her probability of success on the third call is only 0.1.
What is the probability that she makes the sale to a given customer ?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p(sale on first call) = .3 and p(no sale on first call) = .7
p(sale on second call) = .2 and p(no sale on second call) = .8
p(sales on third call) = .1 and p(no sale on third call) = .9
she can make the sale on the first call, or on the second call, or on the third call.
the probability she makes the sale on the first call is .3.
the probability she makes the sale on the second call is .7 * .2.
that is the probability she doesn't make the sale on the first call * the probability that she does make the sale on the second call.
the probability she makes the sale on the third call is .7 * .8 * .1.
that is the probability she doesn't make the sale on the first call * the probability she doesn't make the sale on the second call * the probability she makes the sale on the third call.
her probability of making a sale on either the first call or the second call or on the third call is equal to .3 + (.7 * .2) + (.7 * .8 * .1).
this would be equal to .496.
that's what i think your answer will be, if i am correct.
|
|
|
