Question

Of the cartons produced by a​ company, 66​% have a​ puncture, 44​% have a smashed​ corner, and 0.50.5​% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner.

Answer 1

Answer:

0.095 or 9.5%

Step-by-step explanation:

Puncture → P(P) = 0.06

Smashed corner → P(S) = 0.04

Puncture and smashed corner → P(P and S) = 0.005

The probability that a randomly selected carton has a puncture or a smashed corner is given by the probability of a puncture, added to the probability of a smashed corner, subtracted by the probability of both:

$P(P\ or\ S) =P(P)+P(S) - P(P\ and\ S)\\P(P\ or\ S) =0.06+0.04-0.005\\P(P\ or\ S) =0.095=9.5\%$

The probability is 0.095 or 9.5%.