Question

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.)

Answer 1

Full Question

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.)

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

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