1. ## Probability

This problem is probably really easy. I just cant seem to get it.

Untrained dog has 70% probability of detecting intruder, and given detection, a 50% chance of scaring the intruder away. what is the probability an untrained house dog successfully scares a burglar?

Im guessing it uses Bayes' Th'm.
So, P(A|B)= P(A) P(B|A) / P(B)
P(untrained dog scares intruder)= P(A)=?
P(untrained dog detects intruder)= P(B)= .7
P(given detection untrained dog scares intruder)= P(A|B)= .50
I need to solve for P(A).
P(A) = P(A|B) P(B) / P(B|A)

I dont have P(B|A) either. What am I missing?

2. Originally Posted by PensFan10
This problem is probably really easy. I just cant seem to get it.

Untrained dog has 70% probability of detecting intruder, and given detection, a 50% chance of scaring the intruder away. what is the probability an untrained house dog successfully scares a burglar?

Im guessing it uses Bayes' Th'm.
So, P(A|B)= P(A) P(B|A) / P(B)
P(untrained dog scares intruder)= P(A)=?
P(untrained dog detects intruder)= P(B)= .7
P(given detection untrained dog scares intruder)= P(A|B)= .50
I need to solve for P(A).
P(A) = P(A|B) P(B) / P(B|A)

I dont have P(B|A) either. What am I missing?
Let D represent the event dog detects intruder and S represent the event dog scares intruder.

You're given Pr(D) = 0.7 and Pr(S | D) = 0.5. And I'd asssume that Pr(S | D') = 0 ..... So:

Pr(S) = Pr(S | D) Pr(D) + Pr(S | D') Pr(D') = ....

By the way: A tree diagram makes things clear .....