Answer to Question #308120 in Statistics and Probability for peter

Question

On average 20% of the calculators produced in a factory are faulty. Samples of 10 calculators are to be selected at a random each day. Each calculator will be selected and replaced in the set of calculators which have been produced on that day. Calculate the probability that in any one sample, two calculators or less will be faulty. 

Answer 1

Let X be a random variable representing the number of faulty calculators out of 10, then X ~ BIN(10, 0.2)

$P(X≤2)=P(X=0)+P(X=1)+P(X=2)={10 \choose 0}*0.2^0*0.8^{10}+{10 \choose 1}*0.2^1*0.8^9+{10 \choose 2}*0.2^2*0.8^8\approx0.6778$