Respuesta :
COMPLETE QUESTION:
Chris went on a vacation for a week and asked his brother Paul to feed his old cat Charlie. But Paul is forgetful, and Chris is 70% sure Paul will forget to feed his cat. Without food, Charlie will die with probability 0.5. With food, he will die with probability 0.03. Chris came back from vacation and found Charlie alive. What is the probability that Paul forgot to feed Charlie (round off to third decimal place)?
Answer:
The probability that Paul forgot to feed charlie is 0.546
Step-by-step explanation:
Lets denote F the event 'Paul forgot to feed Charlie', and L the even 'Charlie is alive', we have
P(F) = 0.7
P(L|F) = 1-0.5 = 0.5
P(L|F^c) = 1-P(L^c|F^c) = 1-0.03 = 0.97
We want to calculate P(F|L). We will use Bayes formula at the start and the theorem of total probability to calculate P(L).
[tex]P(F|L) = \frac{P(L|F)*P(F)}{P(L)} = \frac{P(L|F)*P(F)}{P(L|F)*P(F)+P(L|F^c)*P(F^c)} \\= \frac{0.5*0.7}{0.5*0.7+0.97*0.3} = \frac{0.35}{0.35+0.291} = 0.546[/tex]
Given that Charlie is alive, the probability that Paul forgot to feed charlie is 0.546.
Answer:
P = 0.546
Step-by-step explanation:
Hi,
This is a question of conditional probability, which means to find probability of a situation given that another event has already occured:
[tex]P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{P(A \cap B)}{P(B)}[/tex]
In this question, we need to find the probability of Charlie being alive if not fed, with the data given below:
[tex]P(Paul\ forgets)= 0.70\\P(Paul\ feeds) = 0.30\\P( Charlie\ dies\ given\ that\ Paul\ forgets) = 0.50\\P( Charlie\ dies\ given\ that\ Paul\ feeds) = 0.03\\[/tex]
From this data, we can infer the following:
The probability of Charlie staying alive in both cases:
[tex]P(Charlie\ stays\ alive) = (0.97 \times 0.30) + (0.5 \times 0.7)\\P(Charlie\ stays\ alive) = 0.641[/tex]
We need to find the probability when not fed:
[tex]P (Charlie\ alive\ when\ not\ fed) = \frac{P( dies | not fed) \times P(Paul forgets) }{P(Charlie\ stays\ alive)}[/tex]
(Remember this is the variation of the conditional probability formula as per our requirement in this question).
[tex]P(Charlie\ alive\ when\ not\ fed) = \frac{(0.5 \times 0.7)}{0.641} = 0.546[/tex]
Hence, the probability of Charlie being alive when Paul forgets is 0.546.
