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Can Defects A company that makes soda pop cans finds that the

probability of producing a can without a puncture is 0.96, the probability that

a can does not have a smashed edge is 0.93, and the probability that a can

does not have a puncture and does not have a smashed edge is 0.893.

(a) Are the events "selecting a can without a puncture" and "selecting a can

without a smashed edge" mutually exclusive? Explain.

(b) If a quality inspector randomly selects a can, find the probability that the

can does not have a puncture or does not have a smashed edge

Respuesta :

Using probability concepts, it is found that:

a) Since [tex]P(A)P(B) \neq 0[/tex], the events are not mutually exclusive.

b) 0.997 = 99.7% probability that the  can does not have a puncture or does not have a smashed edge.

Item a:

Two events, A and B, are mutually exclusive if:

[tex]P(A \cap B) = 0[/tex]

In this problem:

  • Event A: No puncture, thus, [tex]P(A) = 0.96[/tex].
  • Event B: No smashed edges, thus [tex]P(B) = 0.93[/tex]

The problem also states that the probability of neither is [tex]P(A \cap B) = 0.893[/tex].

Since [tex]P(A)P(B) \neq 0[/tex], the events are not mutually exclusive.

Item b:

Using Venn probabilities, the "or" probability is given by:

[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]

With the given values:

[tex]P(A \cup B) = 0.96 + 0.93 - 0.893 = 0.997[/tex]

0.997 = 99.7% probability that the  can does not have a puncture or does not have a smashed edge.

A similar problem is given at https://brainly.com/question/23855473

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