Question

Hydro Flask, a company that specializes in stainless steel water bottles, offers a lifetime warranty on its products. Bottle A comprised 58% of Hydro Flask’s sales, while Bottle B made up the other 42%. 7.5% of Bottle A sales used the warranty, and 3.2% of Bottle B sales used the warranty.

Given the prior probabilities and the conditional probabilities, what is the joint probability that the bottle is Bottle A and requires the use of the warranty? Specify your result as a proportion with four decimals and no units.

Answer 1

sol:

The Joint probability of two independent events is represented by P(A n B) = P(A) x P(B)

P(A) is the probability of event A occuring

P(B) is the probability of event B occuring

Here in the given problem Bottle A covers 58% of sales and 7.5% of Bottle A sales uses warranty

So P(A) = 58/100 x 7.5/100 = 0.0435

Bottle B  covers 42% of sales and 3.2% of Bottle B sales uses warranty

P(B) = 42/100 x 3.2/100 = 0.0134

Thus P(A n B) is P(A) x P(B) = 0.0435 x 0.0134 =0,0005829 when rounded of to 4 decimal places it is 0.0006 x 100 = 0.06%

Thus P(A n B) is 0.06%