Probability ! it is estimated that 0.50% of a large consignment of eggs in a certain supermarket is broken. what is the probability that a customer who randomly selects a dozen of these eggs receives at least one broken egg?

This is binomial prob.  The number of data points, n, is 12; the chances that an egg will be broken are 0.5, and we want to know the prob. that at least 1 egg is broken.  Note that this is the opposite of finding the prob. that none are broken.

Find binompdf(12, 0.5, 0) and sub tract the result from 1.00 to find the desired

binompdf(12,0.5,0) = 2.44*10^(-4)

So the prob. that the customer receives at least 1 broken egg is

1.00000 - 2.44*10^(-4), or 0.9998.  Big probability!  Buy your eggs elsewhere.

The answer for this is 9/1253decimal form 0.007

21, i'm pretty sure it's right lol