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Elanso [62]
3 years ago
9

Of the cartons produced by a​ company, ​% have a​ puncture, ​% have a smashed​ corner, and ​% 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 nothing​%. ​(Type an integer or a decimal. Do not​ round.)
Mathematics
1 answer:
LuckyWell [14K]3 years ago
3 0

Full Question

<em>Of the cartons produced by a​ company, 10​% have a​ puncture, 6​% have a smashed​ corner, and 0.4​% 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 nothing​ ____%. ​(Type an integer or a decimal. Do not​ round.)</em>

<em />

Answer:

P(Punctured\ or\ Smashed\ Corner) = 0.156

Step-by-step explanation:

Given

Puncture\ Corner = 10\%

Smashed\ Corner = 6\%

Punctured\ and\ Smashed\ Corner = 0.4\%

Required

P(Punctured\ or\ Smashed\ Corner)

For non-mutually exclusive event described above, P(Punctured or Smashed Corner) can be calculated as thus;

P(Punctured\ or\ Smashed\ Corner) = P(Punctured\ Corner) + P(Smashed\ Corner) - P(Punctured\ and\ Smashed\ Corner)

Substitute:

10% for P(Puncture Corner),

6% for P(Smashed Corner) and

0.4% for P(Punctured and Smashed Corner)

P(Punctured\ or\ Smashed\ Corner) = 10\% + 6\% - 0.4\%

P(Punctured\ or\ Smashed\ Corner) = 15.6\%

Convert % to fraction

P(Punctured\ or\ Smashed\ Corner) = \frac{15.6}{100}

Convert to decimal

P(Punctured\ or\ Smashed\ Corner) = 0.156

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Hope this Helps!!!

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