1. ## Zany probablity problem.....

A lady had a biopsy on a possibly cancerous tumor. She instructs the doctor to flip a coin. If it comes up heads, will call if the news is good but not if it is bad. If it comes up tails, the doctor will not call no matter what the results are. Let a be the probability that the tumor is cancerous and b the conditional probability that the tumor is cancerous given the doctor does not call.

Find b in terms of a and determine which probability is larger.

Intuition tells me the probability of b is the larger of the two, but I don't know how to show this.

2. Nevermind I think I got it:

I believe B to be bigger and after some calculations I got

B= 2a/(1+a)....can anyone just confirm or reject this answer?

3. Hello, zhupolongjoe!

It took a couple of readings to get the informaton digested.

She instructs the doctor to flip a coin.
If it comes up Heads, he will call if the news is good, but not if it is bad.
If it comes up Tails, the doctor will not call no matter what the results are.

Let $\displaystyle a$ be the probability that the tumor is cancerous
and $\displaystyle b$ the conditional probability that the tumor is cancerous, given the doctor does not call.

Find $\displaystyle b$ in terms of $\displaystyle a$ and determine which probability is larger.

We have: .$\displaystyle \begin{array}{c}P(\text{cancer}) \:=\:a \\ P(\text{cancer }|\text{ not call}) \:=\:b \end{array}$

Consider the probability that he does not call.
There are two cases:

(1) The coin turns up Heads and the tumor is cancerous.
. . .This probability is: .$\displaystyle \tfrac{1}{2}a$

(2) The coin turns up Tails, so he does not call.
. . .This probability is: .$\displaystyle \tfrac{1}{2}$

Hence: .$\displaystyle P(\text{not call}) \:=\:\tfrac{1}{2}a + \tfrac{1}{2}$

From Bayes' Theorem, we have: .$\displaystyle P(\text{cancer }|\text{ not call}) \;=\;\frac{P(\text{cancer }\wedge\text{ not call})}{P(\text{not call})}$

This becomes: .$\displaystyle b \;=\;\frac{\frac{1}{2}a}{\frac{1}{2}a + \frac{1}{2}} \quad\Rightarrow\quad b \:=\:\frac{a}{a+1}$

$\displaystyle \text{We have: }\;b \;=\;a\underbrace{\left(\frac{1}{a+1}\right)}_{\te xt{less than 1}}\quad\hdots \text{ That is, }b\text{ equals }a\text{ times a number less than 1.}$

Therefore: .$\displaystyle b \:<\:a$

4. Hmmm...that's almost what I have....

I have