# Stat and prob help

• Apr 21st 2007, 04:18 PM
schinb64
Stat and prob help
I have three questions that I am not sure of: Help on any of them is greatly appreicated

1) Assume the probability that you will make a sale on a given phone call is .19
a) Find the probability that you will not make a sale on the first three calls ie. your sale will be made after the first three calls
b) Find the probability that your fourth sale is made on the eighth call

2) Among 90 cities that professional group is considering for its next three annual conventions, 36 are in the western part of the united states use the binomial approx to the hypergeometric distribution to find the probability that at most one of the three conventions will be held in the western part of the US

3)According to an airline report roughly 1 piece of luggage out of every 200 that are checked is lost. Suppose that a freq flyer will check 120 bags over a year approx the probabilty that she will lose 2 or more pieces of luggage
• Apr 22nd 2007, 06:52 AM
CaptainBlack
Quote:

Originally Posted by schinb64
I have three questions that I am not sure of: Help on any of them is greatly appreicated

1) Assume the probability that you will make a sale on a given phone call is .19
a) Find the probability that you will not make a sale on the first three calls ie. your sale will be made after the first three calls
b) Find the probability that your fourth sale is made on the eighth call

a)The probability that no sale is made on the first three calls is the product
of the prob that no sale is made on the first and the prob that no sale is
made on the secons and the prob that no sale is made on the third.

So the reqired prob is (1-0.19)^3 ~= 0.531

b) The probability that the fouth sale is made on the eighth call is the
probability of 3 sales on the first seven calls and a sale on the eighth.

So the reqired prob is b(3;7,0.19) 0.19 = 7!/[4! 3!] 0.19^3 (1-0.19)^4 0.19 ~= 0.0196.

(as the nu,ber of sales in N calls has a binomial distribution B(7,0.19)

RonL
• Apr 22nd 2007, 06:56 AM
CaptainBlack
Quote:

Originally Posted by schinb64
2) Among 90 cities that professional group is considering for its next three annual conventions, 36 are in the western part of the united states use the binomial approx to the hypergeometric distribution to find the probability that at most one of the three conventions will be held in the western part of the US

here we are required to treat the number of conference held in the west
over the next three years to have binomial distribution B(3, 36/90)

That at most 1 is held in the west is the same thing as 0 or 1, so the
required probability is:

P = b(0; 3, 0.4) + b(1, 3, 0.4) = 0.6^3 + 3 (0.4) (0.6^2)

RonL