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RUDIKE [14]
3 years ago
6

For a recent evening at a small, old-fashioned movie theater, 35% of the moviegoers were female and 65% were male. There were tw

o 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?
Mathematics
1 answer:
Taya2010 [7]3 years ago
4 0

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

Step-by-step 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)=\frac{P(R^{-}\ and\ F)}{P(F)}

P(R^{-}|F)=\frac{P(R^{-})\ *\ P(F)}{P(F)}

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

P(R^{-}|F)=\frac{0.40 * 0.35}{0.35}

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

<em>This implies that both events are independent</em>

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 <em>--- Given</em>

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