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.

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:

[tex]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\%[/tex]

The probability is 0.095 or 9.5%.

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.

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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.