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Chapter 6: Problem 96

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
The results obtained are a) \(P(E^C) = 0.40\), b) \(P(E \cup F) = 0.75\), and c) \(P(E^C \cap F^C) = 0.34\).

Step by step solution

01

Calculate the Complement of E

The complement of an event E, represented as \(E^C\), is the event that E does not occur. Therefore, to calculate \(P(E^C)\), subtract the probability of E from 1, i.e., \(P(E^C) = 1 - P(E) = 1 - 0.60 = 0.40\).
02

Calculate the Union of E and F

The union of events E and F, represented as \(E \cup F\), is the event that either E or F or both occur. According to the rule of addition, we calculate \(P(E \cup F) = P(E) + P(F)\) since E and F are mutually exclusive events. Hence, \(P(E \cup F) = 0.60 + 0.15 = 0.75\).
03

Calculate the Complement of Both E and F

The complement of both events E and F, represented as \(E^C \cap F^C\), is the event that neither E nor F occur. From the rule of multiplication for the intersection of two independent events, we calculate \(P(E^C \cap F^C)\) as \(P(E^C \cap F^C) = P(E^C) * P(F^C) = 0.40 * 0.85 = 0.34\). The complement of F, \(F^C\), is computed the same way as the complement of E, \(E^C\), i.e., \(P(F^C) = 1 - P(F) = 1 - 0.15 = 0.85\).

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Most popular questions from this chapter

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