Question

It is known that 10% of the calculators shipped from a particular factory are defective. what is the probability that at least one in a random sample of four calculators is defective

Answer 1

P = 10% = 0.1
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)

The brobability of a binomial distribution is given by
$P(x)=^nC_xp^xq^{n-x}$
where: n = 4
$P(0)=^4C_0(0.1)^0(0.9)^4=1\times1\times0.6561=0.6561$
Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439