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For a recent evening at a small, old-fashioned movie theater, 35% of the moviegoers were female and 65% were male. There were two movies playing that evening. One was a romantic comedy, and the other was a World War II film. As might be expected, among the females the romantic comedy was more popular than the war film: 70% of the females attended the romantic comedy. Among the male moviegoers, the romantic comedy also was more popular: 60% of the males attended the romantic comedy. No moviegoer attended both movies. Let F denote the event that a randomly chosen moviegoer (at the small theater that evening) was female and ¯F denote the event that a randomly chosen moviegoer was male. Let R denote the event that a randomly chosen moviegoer attended the romantic comedy and ¯R denote the event that a randomly chosen moviegoer attended the war film. Fill in the probabilities below, and then answer the question that follows. Do not round any of your responses

P(F)=0.35
P(F and R)=
P(F and ¯R)=
P(¯R|F)=
P(R|¯F)=0.60
P(¯F and R)=
P(¯F and ¯R)=
P(¯F)=
What is the probability that a randomly chosen moviegoer attended the romantic comedy?

1 Answer

7 votes

Answer:


P(F\ and\ R) = 0.21


P(F\ and\ R^(-)) = 0.14


P(R^(-)|F)= 0.40


P(F^(-) and\ R) = 0.39


P(F^(-)\ and\ R^(-)) =0.26


P(F^(-)) =0.65

Explanation:

The variables have been defined in the question as:


F = Female\ Moviegoer


F^(') = Male\ Moviegoer


R = Romantic\ Comedy


R^(') = War\ File

Also, we have the following given parameters:


P(F) = 0.35


P(F^(-)) =0.65


P(R) =0.60


P(R^(-)) = 0.40

The solution is as follows:


a.\ P(F\ and\ R)


P(F\ and\ R) = P(F) * P(R)

Substitute values for P(F) and P(R)


P(F\ and\ R) = 0.35 * 0.60


P(F\ and\ R) = 0.21


b.\ P(F\ and\ R^(-))


P(F\ and\ R^(-)) = P(F) * P(R^(-))

Substitute values for P(F) and P(R-)


P(F\ and\ R^(-)) = 0.35 * 0.40


P(F\ and\ R^(-)) = 0.14


c.\ P(R^(-)|F)


P(R^(-)|F)=(P(R^(-)\ and\ F))/(P(F))


P(R^(-)|F)=(P(R^(-))\ *\ P(F))/(P(F))

Substitute values for P(F) and P(R-)


P(R^(-)|F)=(0.40 * 0.35)/(0.35)


P(R^(-)|F)= 0.40

This implies that both events are independent


d.\ P(F^(-) and\ R)


P(F^(-) and\ R) = P(F^(-)) * P(R)

Substitute values for P(F-) and P(R)


P(F^(-) and\ R) = 0.65 * 0.60


P(F^(-) and\ R) = 0.39


e.\ P(F^(-)\ and\ R^(-))


P(F^(-)\ and\ R^(-)) =P(F^(-)) * P(R^(-))

Substitute values for P(F-) and P(R-)


P(F^(-)\ and\ R^(-)) =0.65 * 0.40


P(F^(-)\ and\ R^(-)) =0.26


f.\ P(F^(-))


P(F^(-)) =0.65 --- Given

User Erich Schreiner
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