Question 1117767
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.