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?

Answer:

[tex]P(F\ and\ R) = 0.21[/tex]

[tex]P(F\ and\ R^{-}) = 0.14[/tex]

[tex]P(R^{-}|F)= 0.40[/tex]

[tex]P(F^{-} and\ R) = 0.39[/tex]

[tex]P(F^{-}\ and\ R^{-}) =0.26[/tex]

[tex]P(F^{-}) =0.65[/tex]

Step-by-step explanation:

The variables have been defined in the question as:

[tex]F = Female\ Moviegoer[/tex]

[tex]F^{'} = Male\ Moviegoer[/tex]

[tex]R = Romantic\ Comedy[/tex]

[tex]R^{'} = War\ File[/tex]

Also, we have the following given parameters:

[tex]P(F) = 0.35[/tex]

[tex]P(F^{-}) =0.65[/tex]

[tex]P(R) =0.60[/tex]

[tex]P(R^{-}) = 0.40[/tex]

The solution is as follows:

[tex]a.\ P(F\ and\ R)[/tex]

[tex]P(F\ and\ R) = P(F) * P(R)[/tex]

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

[tex]P(F\ and\ R) = 0.35 * 0.60[/tex]

[tex]P(F\ and\ R) = 0.21[/tex]

[tex]b.\ P(F\ and\ R^{-})[/tex]

[tex]P(F\ and\ R^{-}) = P(F) * P(R^{-})[/tex]

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

[tex]P(F\ and\ R^{-}) = 0.35 * 0.40[/tex]

[tex]P(F\ and\ R^{-}) = 0.14[/tex]

[tex]c.\ P(R^{-}|F)[/tex]

[tex]P(R^{-}|F)=\frac{P(R^{-}\ and\ F)}{P(F)}[/tex]

[tex]P(R^{-}|F)=\frac{P(R^{-})\ *\ P(F)}{P(F)}[/tex]

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

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

[tex]P(R^{-}|F)= 0.40[/tex]

This implies that both events are independent

[tex]d.\ P(F^{-} and\ R)[/tex]

[tex]P(F^{-} and\ R) = P(F^{-}) * P(R)[/tex]

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

[tex]P(F^{-} and\ R) = 0.65 * 0.60[/tex]

[tex]P(F^{-} and\ R) = 0.39[/tex]

[tex]e.\ P(F^{-}\ and\ R^{-})[/tex]

[tex]P(F^{-}\ and\ R^{-}) =P(F^{-}) * P(R^{-})[/tex]

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

[tex]P(F^{-}\ and\ R^{-}) =0.65 * 0.40[/tex]

[tex]P(F^{-}\ and\ R^{-}) =0.26[/tex]

[tex]f.\ P(F^{-})[/tex]

[tex]P(F^{-}) =0.65[/tex] --- Given

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