LUANDATO wrote:In a certain factory, the production line which produces the bags has the probability that a bag selected at random is defective is 0.02. If 6 bags are selected at random, what is the probability that at least one bag is defective?

$$A.\ 0.02^6$$
$$B.\ \left(0.02\right)\left(0.98\right)^6$$
$$C.\ \ 1-\left(0.02\right)^6$$
$$D.\ \ 0.98^6$$
$$E.\ \ 1-0.98^6$$

We want P(select at least 1 defective bag)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 defective bag) = 1 - P(not getting at least 1 defective bag)
What does it mean to not get at least 1 defective bag? It means getting zero defective bags.
So, we can write: P(getting at least 1 defective bag) = 1 - P(getting zero defective bags)

ASIDE: If P(bag is defective) = 0.02, then P(bag is NOT defective) = 0.98

P(getting zero defective bags)
P(getting zero defective bags) = P(1st bag is good AND 2nd bag is good AND 3rd bag is good AND 4th bag is good AND 5th bag is good AND 6th bag is good)
= P(1st bag is good) x P(2nd bag is good) x P(3rd bag is good) x P(4th bag is good) x P(5th bag is good) x P(6th bag is good)\
= 0.98 x 0.98 x 0.98 x 0.98 x 0.98 x 0.98
= 0.98^6

So, P(select at least 1 defective bag) = 1 - 0.98^6
= E

Cheers,
Brent