If 1.5% of the antennas on new Nokia cell phones are defective, then
for a random sample of 10 antennas find the probability that:
1) None of the antennas are defective
2) 3 or more of the antennas are defective
3) At most 3 are defective
4) All 10 antennas are defective

Hi!!

Here we mus use the Poisson distribution, for references see:
"Poisson distribution" from Wikipedia:
http://www.wikipedia.org/wiki/Poisson_distribution

For Poisson distribution with expected value (Poisson parameter) h we
have that:
          e^-h * h^k  
P(X=k) = ------------
              k!

where:
e is the base of the natural logarithm (e = 2.71828...), 
k is an integer number, 
h is a positive real number, equal to the expected number of
occurrences that occur during the given interval (in this case h =
10*0.015 =0.15).

To answer this question just plug the values in the formula showed
above and use a calculator or simply use the following online
calculator (note that h is m).
http://www.changbioscience.com/stat/prob.html


1) None of the antennas are defective:

P(X = 0) = 0.860708
 


2) 3 or more of the antennas are defective:

P(X >= 3) = 1 - P(X<3) =
          = 1 - P(X=0) - P(X=1) - P(X=2) = 
          = 5.028624 * 10^-4
  


3) At most 3 are defective:

P(X =< 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0.9999813



4) All 10 antennas are defective:

P(X = 10) = 1.36775 * 10^-15



I hope this helps you. Please do not hesitate to request for a
clarification if you find something unclear.

Regards.
livioflores-ga

Thanks for the calculator link, now I can use it myself!

Some comments:

1. Apparently livioflores plugged 15% in it instead of 1.5%

2. This is not a Poisson, but Binomial distribution, so when you use
that calculator make sure you choose "Binomial" with p=0.015 and N=10.