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soldier1979 [14.2K]
3 years ago
15

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.
Mathematics
1 answer:
Xelga [282]3 years ago
6 0

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%.

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