Probabiltiy Question

• Aug 26th 2009, 01:36 PM
doolindalton
Probabiltiy Question
I think I know the answer but would like to hear the forum's answer. What was the logic was in getting at the answer? Also, I copy pasted this from an actual exam so no need to wonder if there was a typo or anything.

The Question:

The probability that a visit to a primary care physician’s (PCP) office results in neither lab work nor referral to a specialist is 35%. Of those coming to a PCP’s office, 30% are referred to specialists and 40% require lab work. Determine the probability that a visit to a PCP’s office results in both lab work and referral to a specialist.

(A) 0.05
(B) 0.12
(C) 0.18
(D) 0.25
(E) 0.35
• Aug 26th 2009, 03:37 PM
pickslides
There are a few ways to tackle ths problem. You are given that,

$P(A'\cup B') = 0.35$

Where

$P(B) = 0.3$

$P(A) = 0.4$

You are required to find $P(A\cap B)$

If $P(A'\cup B') = 0.35 \Rightarrow P(A\cup B) = 0.65$

Now as $P(A\cup B) = P(A)+P(B)- P(A\cap B)$

$0.35 = 0.4+0.3- P(A\cap B)$

$P(A\cap B) = 0.05$
• Aug 26th 2009, 04:14 PM
doolindalton
Thank you. That is the answer I got and you just won me lunch from my stubborn friend. And thanks for showing the proof. I was trying to explain the logic in words and was having a hard time getting through.

Several people I showed this to, including my friend, picked .12 (30% * 40%) because they were thinking,

P(A) = 30%
P(B) = 40%

So, P(A and B) = P(A) * P(B)

Obviously, 0.12 was the trap answer for those not thinking it through.
By the way, is this formula correct? If so, why doesn't it work in this problem?
• Aug 26th 2009, 04:50 PM
pickslides
Quote:

Originally Posted by doolindalton
Thank you. That is the answer I got and you just won me lunch from my stubborn friend.

Great! I love a free lunch.

Quote:

Originally Posted by doolindalton

So, P(A and B) = P(A) * P(B)

Obviously, 0.12 was the trap answer for those not thinking it through.
By the way, is this formula correct? If so, why doesn't it work in this problem?

This does work in the case where A is independant of B. In your case A and B were not independant of each other.
• Aug 27th 2009, 06:23 AM
doolindalton
$0.35 = 0.4+0.3- P(A\cap B)$

$P(A\cap B) = 0.05$

Minor correction to the above.
You meant to say

0.65 = 0.4 + 0.3 - X
=> X = .05