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I have been working on this question for a while and I haven't obtained any reasonable results:

In a city, 70% of the inhabitants are non-smokers. Specialists estimate that there is a 45% chance that smokers will suffer from lung cancer at some point their lives while the probability is 10% for non-smokers. If a person is chosen at random in this city, what is the probability that this person will not develop lung cancer given that this person is a non-smoker?

I know that:

P(B|A) = P(B ^ A) / P(A)

Where the symbol ^ indicates intersection. To this question, I believe, this formula applies as:

P(will not dev. lung cancer|non-smoker) = P(no lung cancer ^ non-smoker) / P(non-smoker)

It is given that P(non-smoker) is 0.7.

However, obtaining the intersection of no lung cancer and non-smoker is the problem for me; I create a Venn diagram such that A (no-lung cancer) and B (non-smoker). Yet, how do I calculate the intersection of the two?

Is there something that I am missing because of the word play in the question?

arin
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2 Answers2

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Much of the information in the question is superfluous. All you need to know is that the probability that a non-smoker will develop lung cancer at some point is $0.1$. Thus, the probability that a non-smoker will not develop lung cancer at some point is $1-0.1=0.9$, or $90$% if you prefer to express your probabilities as percentages.

The $70$% figure is irrelevant: it doesn’t matter how likely a randomly selected person is to be a non-smoker, because we’re told that in fact a non-smoker was chosen. For the same reason, probabilities involving smokers are irrelevant.

Brian M. Scott
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  • Thanks for your answer. I believe the same applies for P(is a smoker and will develop lung cancer) = 45% and P(will not develop lung cancer|is smoker) = 55%, since this information is already present in the info given within the question? – arin Apr 02 '13 at 02:29
  • @arin: Your second probability is right, but the first isn’t: $$\Bbb P(\text{will develop lung cancer}\mid\text{smoker})=0.45;,$$ but $$\Bbb P(\text{smoker and will develop lung cancer})=0.3\cdot0.45=0.135;.$$ – Brian M. Scott Apr 02 '13 at 02:33
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Let the events be:

  • $S$: Choose a smoker at random
  • $C|S$: Develop cancer, given is a smoker
  • $C|\hat S$: Develop cancer, given is not a smoker

Now, from the problem statement we know:

  • $P(S)$= 0.7
  • $P(C|S)$=0.45
  • $P(C|\hat S)$=0.10

The answer to the question as you stated it is $1-P(C| \hat S) = 1-0.10 = 0.90$

But this does not utilize all the information given in the problem statement, in most textbooks they want to show a typical example of the law of total probability that may be more illustrative, like obtaining the probability of develping cancer from a person chosen at random. That would be like this:

From total probability law, note that $C$ and $\hat C$ are partitions of the probability space (this is, they are pairwise disjoint subsets of the probability space):

  • $P( \hat C)= 1 - P(C) = 1 - (P(C,S) + P(C, \hat S))$

And using the conditional probability definition that you mentioned, we obtain:

  • $P(C,S) = P(C|S)*P(S) = 0.45 * 0.7 = 0.315$
  • $P(C, \hat S) = P(C| \hat S)*P(\hat S) = 0.10 * 0.3 = 0.03$

Hence:

  • $P(\hat C) = 1 - (0.315 + 0.03) = 0.655$
papirrin
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