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Chapter 6: Problem 90

In a school machine shop, \(60 \%\) of all machine breakdowns occur on lathes and \(15 \%\) occur on drill presses. Let \(E\) denote the event that the next machine breakdown is on a lathe, and let \(F\) denote the event that a drill press is the next machine to break down. With \(P(E)=.60\) and \(P(F)=.15\), calculate: a. \(P\left(E^{C}\right)\) b. \(P(E \cup F)\) c. \(P\left(E^{C} \cap F^{C}\right)\)

Short Answer

Expert verified
a. The probability of the complement of E is 0.40. b. The probability of the union of E and F is 0.75. c. The probability of intersection of the complements of E and F is 0.25.

Step by step solution

01

Calculate the Complement of E

The complement of an event refers to the scenario where the event does not occur. The probability of the complement of E, denoted by \(P(E^{C})\), is given by the formula \(P(E^{C}) = 1 - P(E)\). Now substitute \(P(E) = 0.60\) into the equation to get \(P(E^{C}) = 1 - 0.60 = 0.40\)
02

Calculate the Union of E and F

The union of two events refers to the probability that at least one of the events happens. The probability of the union of E and F, denoted by \(P(E \cup F)\), is provided by the formula \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). Here, it is given that Events E and F are mutually exclusive (cannot occur at the same time). Hence, \(P(E \cap F) = 0\). Substituting the given values into the formula, we get \(P(E \cup F) = 0.60 + 0.15 - 0 = 0.75\)
03

Calculate the Intersection of Complements of E and F

The intersection of the complements of two events refers to the scenario where neither of the events occur. The probability of the intersection of the complements of E and F, denoted by \(P(E^{C} \cap F^{C})\), is given by the formula \(P(E^{C} \cap F^{C}) = 1 - P(E \cup F)\). Substituting the value of \(P(E \cup F)\) from Step 2 into the formula, we get \(P(E^{C} \cap F^{C}) = 1 - 0.75 = 0.25\)

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Most popular questions from this chapter

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