1. ## random variables

A store has found that 10% of the items it sells are retruend after Christmas. On one day, a total of 50 items are sold by 5 clerks: Clerk A sold 5 items, Clerk B sold 20 items, and the rest were sold by the other 3 clerks.

a) what is the probability that none of the items sold by Clerk A will be returned?
ans: 0.5905

b) what is the probability that exactly 3 of the items sold by Clerk B will be returned?
ans: 0.190

c) what is the probability that between 2 and 7 (inclusively) of the items sold by the other 3 clerks will be returned?
ans: 0.727

d) what is the probability that between 3 and 6 (inclusively) of the items sold by Clerk B and that exactly 4 of the items sold by the other 3 clerks will be returned?
ans: 0.044

2. Originally Posted by skhan
A store has found that 10% of the items it sells are retruend after Christmas. On one day, a total of 50 items are sold by 5 clerks: Clerk A sold 5 items, Clerk B sold 20 items, and the rest were sold by the other 3 clerks.

a) what is the probability that none of the items sold by Clerk A will be returned?
ans: 0.5905
This used binomial probability.
$\displaystyle {n \choose m}p^m(1-p)^{n-m}$
Over here you need all 5 not to. The probability of that not happening is 90% thus,
$\displaystyle {5 \choose 5}(.9)^5(.1)^0=.5905$

Originally Posted by skhan
b) what is the probability that exactly 3 of the items sold by Clerk B will be returned?
ans: 0.190
$\displaystyle {20 \choose 3}(.1)^3(.9)^{17}\approx .1901$

Originally Posted by skhan
c) what is the probability that between 2 and 7 (inclusively) of the items sold by the other 3 clerks will be returned?
ans: 0.727
Here you have 25 of them, thus you need to find,
$\displaystyle \sum ^7_ {k=2} {25 \choose k }(.1)^k(.9)^{25-k}\approx .7265$
Originally Posted by skhan
d) what is the probability that between 3 and 6 (inclusively) of the items sold by Clerk B and that exactly 4 of the items sold by the other 3 clerks will be returned?
ans: 0.044
This is probability of two events. First find between 3 and 6 which is,
$\displaystyle \sum^6_{k=3}{20\choose k}(.1)^k(.9)^{20-k}\approx .3207$
And then probability exactly 4 which is,
$\displaystyle {25 \choose 4}(.1)^4(.9)^{25-4}\approx .1384$
Now multiply them to get,
$\displaystyle \approx .04438$.

Credits
--------
CaptainBlack for the nice trick for combinations using LaTeX really helped.

3. wow thanks!