Question

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 1

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-)

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

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

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