Respuesta :
Answer:
The probability that Bill ate turnips for dinner given that he gets sick is 0.6667.
Step-by-step explanation:
Denote the events as follows:
T = Bill eats turnips
E = Bill eats eggplant
A = Bill eats asparagus
X = Bill gets sick.
Given:
P (X | T) = 0.50
P (X | E) = 0.20
P (X | A) = 0.05
P (T) = P (E) = P (A) = 0.33
Compute the value of P (X) using the law of total probability as follows:
[tex]P(X)=P(X|T)P(T)+P(X|E)P(E)+P(X|A)P(A)\\=(0.50\times0.33)+(0.20\times0.33)+(0.05\times0.33)\\=0.2475[/tex]
The probability that Bill gets sick is 0.2475.
Compute the probability that Bill ate turnips for dinner given that he gets sick as follows:
[tex]P(T|X)=\frac{P(X|T)P(T)}{P(X)}=\frac{0.50\times0.33}{0.2475}=0.6667[/tex]
Thus, the probability that Bill ate turnips for dinner given that he gets sick is 0.6667.
Answer:
P(T/S) = 0.667 .
Step-by-step explanation:
We are given that three different vegetables, turnips, eggplant and asparagus, make Bill sick sometimes.
Let Probability of Bill eating turnips = P(T) = 1/3
Probability of Bill eating eggplant = P(E) = 1/3
Probability of Bill eating asparagus = P(A) = 1/3
Also, let S = event that Bill is sick after any of these three vegetables
So, Probability of Bill becoming sick if he eats turnips = P(S/T) = 0.50
Probability of Bill becoming sick if he eats eggplant = P(S/E) = 0.20
Probability of Bill becoming sick if he eats asparagus = P(S/A) = 0.05
Now, we are given that Bill gets sick after dinner where he ate one of those three vegetables, probability that he ate turnips = P(T/S)
Using Bayes' Theorem we get;
P(T/S) = [tex]\frac{P(T)*P(S/T)}{P(T)*P(S/T) + P(E)*P(S/E) + P(A)*P(S/A)}[/tex]
= [tex]\frac{0.5*(1/3)}{0.5*(1/3) + 0.2*(1/3) + 0.05*(1/3)}[/tex] = [tex]\frac{0.5*(1/3)}{(1/3)*[0.5+0.2+0.05]}[/tex] = [tex]\frac{0.5}{0.75}[/tex] = 0.667
So, required probability is 0.667 .
