Probability in a sales call
Message posted by corin1 (via 205.205.139.34) on October 22, 2001 at 12:11 AM (ET)

I am having a little trouble with the following question:
A door to door salesperson sells rug shampoo in three tube sizes: small, large and giant. The probability of finding a person at home is 0.6. If the salesperson does find someone at home, the probabilities are 0.5 that there will not be a sale, 0.2 that a small tube will be sold, 0.2 that a large tube will be sold and 0.1 that a giant tube will be sold. The probabiltity of selling more than one tube of shampoo at a house is zero.
A: Find the probability that in one call no shampoo will be sold.
B: Find the probability that in one call either a large tube or a small tube will be sold.

I have come up with (.6)(.4)-(.6)(.3)=0.7 for A

And 0.06 + 0.03/ .06 + .06 + .03 + .03 + .04= 0.1059

I did this using a tree, but something tells me I am off by a mile. I would appreciate any help.
Thanks

READERS RESPOND:
(In chronological order. Most recent at the bottom.)

Re: Probability in a sales call
Message posted by Tomi (via 154.32.143.87) on October 22, 2001 at 3:33 PM (ET)

P(noone in) + P(someone in)*P(no sale)

B: .6*(.2+.1)=.18
P(someone in)*P(either large or small)

Re: Probability in a sales call
Message posted by Corin1 (via 205.205.139.97) on October 22, 2001 at 6:07 PM (ET)