A person visits three friends. He found that his wallet was lost when he comes back. The probabilities that his wallet will be lost at each friends’ home are 0.4, 0.2, and 0.1, respectively, in the order of visiting. Given that his wallet was lost at one of his three friend’s homes, what is the probability that his wallet was left at the first friend’s home? Hint: Define W = the wallet was lost at one of the three homes, Ai = the wallet was lost at the ith friend’s home, i=1, 2, 3. The Question gives P(A1)=0.4, P(A2|A1c)=0.2, P(A3|(A1cintersectionA2c))=0.1, and we want to calculate P(A1|W). You can find P(W) by either working on its complement or summing up the probability that all Ai occurs.

0.5714

Step-by-step explanation:

P(1 home) = P(2 home ) = P(3 home) = 1/3

P(Wallet lost | 1 home) = 0.4

P(Wallet lost| 2 home) = 0.2

P(Wallet lost | 3 home) = 0.1 ,

With the use of Bayes' theorem which in probability describes the probability of an event, based on prior knowledge of conditions that might be related to the event.

P(A|B) = P(B|A) P(A) / P(B)

In most cases, you can’t just plug numbers into an equation; Your “tests” and “events” have to be figured out first. For two events, A and B, Bayes’ theorem allows you to figure out p(A|B) (the probability that event A happened, given that test B was positive) from p(B|A) (the probability that test B happened, given that event A happened).

Application of this formula,

P(1 Home | Wallet lost) =

P(1 home) × P(Wallet lost | 1 home) /

[ P(1 home) ×P(Wallet lost | 1 home) + P(2 home) × P(Wallet lost | 2 home)+ P(3 home) × P(Wallet lost | 3 home)]

= ( 1/3 × 0.4) / [ ( 1/3 × 0.4) + ( 1/3 × 0.2) + (1 / 3 × 0.1) ]

= 0.5714

Therefore, the probability that his wallet was left at his first friend home is 0.5714

abby has a greater debt becasue she owea 5 more then alan does