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The probability that Chloe acts hungry at 7pm given that she has already eaten dinner is 0.5. The probability that Chloe acts hungry at 7pm given that she has not had dinner is 0.99. Assume that there is probability 0.9 that Chloe has eaten dinner by 7pm on a given day. a) (1 pt) If Chloe is acting hungry at 7pm, find the probability that she has not had dinner yet. b) (1 pt) If Chloe is not acting hungry at 7pm, find the probability that she has already had dinner.

Respuesta :

Answer:

a) P(  [tex]Y^{C}[/tex] | X ) = 0.180

b) P(Y | [tex]X^{C}[/tex]  ) = 0.998

Step-by-step explanation:

Let

P(X) - Probability that he acts hungry

P(Y) - Probability that he had ate dinner,

Given,

P(X | Y) = 0.5

P(X | [tex]Y^{C}[/tex]  ) = 0.99

P(Y) = 0.9

a.)

P(  [tex]Y^{C}[/tex] | X ) =  [tex]\frac{P( X | Y^{C} ). P(Y^{C} ) }{P( X | Y^{C} ). P(Y^{C} ) + P( X | Y ) . P (Y) }[/tex]

                  = [tex]\frac{(0.99)(0.1)}{(0.99)(0.1) + (0.5)(0.9)} = \frac{0.099}{0.549}[/tex] = 0.180

⇒P(  [tex]Y^{C}[/tex] | X ) = 0.180

b.)

P(Y | [tex]X^{C}[/tex]  ) = [tex]\frac{(1 - P(X | Y ) ) . P(Y)}{P( X^{C} ) } = \frac{(1 - P(X | Y ) ) . P(Y)}{ 1 - P( X ) }[/tex]

                 = [tex]\frac{(1 - 0.5)(0.9)}{1 - [(0.99)(0.1) + (0.5)(0.9) ]} = \frac{(0.5)(0.9)}{1 - [(0.099) + (0.45) ]} = \frac{0.45}{1 - [0.540]} = \frac{0.45}{0.451}[/tex] = 0.998

⇒P(Y | [tex]X^{C}[/tex]  ) = 0.998

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