Ms. Aquina has just had a biopsy on a possibly cancerous tumor. Not wanting to spoil a weekend family event, she does not want to hear any bad news in the next few days

# Ms. Aquina has just had a biopsy on a possibly cancerous tumor. Not wanting to spoil a weekend family event, she does not want to hear any bad news in the next few days

J
240 points

Ms. Aquina has just had a biopsy on a possibly cancerous tumor. Not wanting to spoil a weekend family event, she does not want to hear any bad news in the next few days. But if she tells the doctor to call only if the news is good, then if the doctor does not call, Ms. Aquina can conclude that the news is bad. So, being a student of probability, Ms. Aquina instructs the doctor to flip a coin. If it comes up heads, the doctor is to call if the news is good and not call if the news is bad. If the coin comes up tails, the doctor is not to call. In this way, even if the doctor doesn’t call, the news is not necessarily bad. Let α be the probability that the tumor is cancerous; let β be the conditional probability that the tumor is cancerous given that the doctor does not call.

(a) Which should be larger, α or β?

(b) Find β in terms of α, and prove your answer in part (a).

conditional probability

8.7k points

After the doctor receives the results of the biopsy he will flip a coin. Define the following events,

B is the event of bad news
G is the event of good news
C is the event that the doctor calls

Which should be larger, α or β

let $C^c$ be the event that the doctor does not call. We are told that,

α = P(B) = tumor is cancerous

β = $P(B|C^c)$ = conditional probability tumor is cancerous given that the doctor doesn't call

also

we know that, $P(C^c|B) = 1$ since Ms Aquina requested no call for bad news regardless of whether the coin lands heads or tails.

Also, $P(C^c|G)=\frac{1}{2}$ since the doctor will call if the news is good and the coin lands heads, and will not call under any circumstances if the coin lands tails.

Now IF the doctor does not call, then β > α under the condition that

$$P(C^c|B) \ge P(C^c|G)$$

So we expect β to be larger.

Find β in terms of α, and prove your answer in part (a)

$$\beta = P(B|C^c)=\frac{P(C^c|B)P(B)}{P(C^c)}=\frac{1(\alpha)}{3/4}=\frac{4}{3}\alpha > \alpha$$

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