# Probability with bayes's theorem

• Mar 4th 2013, 03:47 PM
hiensteve
Probability with bayes's theorem

Joe Greedy believes that he has found the location where a Spanish galleon that was transporting a gold shipment worth \$10 million sank. He estimates that the probability that he has correctly determined the general area where the shipwreck took place is p. If the treasure is indeed in this area, the probability that he would find it during a dive is q. Dives are independent. Each dive costs c thousands of dollars and Joe has enough money to fund N dives.
Question: Use Bayes’ theorem to determine p_k, the conditional probability that the treasure be in the area given that k unsuccessful dives have been made previously.
• Mar 4th 2013, 06:56 PM
chiro
Re: Probability with bayes's theorem
Hey hiensteve.

Hint: If you have un-successful dives then this means that D = (D1 = unsuccessful AND D2 = unsuccessful ... AND Dk = Unsuccessful) where you want to find P(Treasure In Area|D).

Remember that for independent events P(A and B) = P(A)P(B).
• Mar 4th 2013, 10:21 PM
hiensteve
Re: Probability with bayes's theorem
Quote:

Originally Posted by chiro
Hey hiensteve.

Hint: If you have un-successful dives then this means that D = (D1 = unsuccessful AND D2 = unsuccessful ... AND Dk = Unsuccessful) where you want to find P(Treasure In Area|D).

Remember that for independent events P(A and B) = P(A)P(B).

T: Treasure in the area
F: Treasure is not in the area
Dk: k dives have been made unsuccessfully

Then

Pk=P(T/Dk)={P(T). P(Dk/T)}/ {P(Dk/T).P(T)+ P(F ).P(Dk/F)}

= {(p.q)^k . (1-q)^k }/ {(p.q)^k . (1-q)^k +(1-p.q)^k.1}

P(Dk/T)= (1-q)^k because here Dk will happen in the case T happen, in other words, it means treasure’s in the area or p =1.
P(Dk/F)= 1 because here Dk will happen in the case T happen, in other words, it means treasure’s not in the area and k dives will be unsuccessful definitely, or q = 0->(1-p.q)^k = 1