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I'm stumped on determining P(ABC) of Part A. My understanding is:

  • Calculate the total number of patients (100)

  • Calculate individual $P(A), P(B),$ and $P(C)$ $(0.4; 0.35; 0.24 $ respectively)

  • Multiply $P(A)P(B)P(C) = ANS (0.0336)$

This answer should be the same as $8/20$ or the population of serious patients, under 40, whose parents had diabetes but this fraction comes out to 0.08--what am I missing here? I appreciate your help.


"What am I missing here?"

$Pr(A\cap B)$ is generally NOT equal to $Pr(A)\times Pr(B)$.

You have that $Pr(A\cap B)=Pr(A)\times Pr(B)$ if and only if $A$ and $B$ are independent events. You may not just break apart intersections inside of a probability as a product of probabilities in general unless you know ahead of time that the events are independent.

Similarly, $Pr(A\cup B)$ is generally NOT equal to $Pr(A)+Pr(B)$. They happen to be equal if and only if $Pr(A\cap B)=0$. It is only in those special circumstances where you can break things apart by multiplication or addition like this.

Instead, in your specific situation, count the total number of cases which satisfy all of the descriptors $A,B,C$ simultaneously and divide by the total number of recorded cases.

  • $\begingroup$ Note that the numbers in the table are given in percentages, not raw numbers, so the answer literally appears in the table with no calculation necessary. $\endgroup$ – Robert Shore Feb 12 at 0:44
  • $\begingroup$ Right, the answer is "8/100" which is the number of patients that satisfy each of the criteria. But, from the method that was described by JMoravitz (Adding 40 Patients from Pr(A) + 35 Patients from Pr(B) + 58 patients from Pr(C) / 100 total patients in the study) I get a probability of 133% which is wrong. Instead, multiplying these probabilities gets me the right answer--right? $\endgroup$ – David King Feb 12 at 0:50
  • $\begingroup$ @DavidKing What? No, I never told you to add $Pr(A)+Pr(B)+Pr(C)$... I was saying to add $15+10+8+2+15+20+20+10$ because at first glance I missed that these were percentages and assumed that we were talking about a limited survey of 100 cases and had not yet checked that the numbers added up to 100. Of course, missing that observation in no way affects the answer. $\endgroup$ – JMoravitz Feb 12 at 1:06
  • $\begingroup$ @DavidKing The answer to $Pr(A\cap B\cap C)$, as pointed out already, is very simply $\frac{8}{100}$ since $8$ is the number in the corresponding part in the chart which corresponds to each of $A,B,C$ all being simultaneously true (and since the sum of all numbers is 100, or equivalently since we were told the numbers in the table correspond to percentages). $\endgroup$ – JMoravitz Feb 12 at 1:12
  • $\begingroup$ @DavidKing As an aside, the scan of the page asked you to find $Pr(B\cap C)$ but in your post you write that you tried to find $Pr(C)$. Neither of these are equal to $0.24$. $\endgroup$ – JMoravitz Feb 12 at 1:14

$P(ABC) = P(A \; \text{and} \; B \; \text{and} \; C) \neq P(A)P(B)P(C)$. The equality holds iff events $A$, $B$, $C$ are independent.

In your case you need to compute the probability that a patient has a serious case (A) AND is below 40 (B) AND his parents are diabetic (C). From your data table, you can see that 8 patients satisfy all 3 criteria. Since the total number of data points you have is 100, your probability estimate $P(ABC) = 8/100 = 0.08$.

  • $\begingroup$ So, in this case, each of the probabilities would be considered independent in this problem? $\endgroup$ – David King Feb 12 at 0:56
  • 1
    $\begingroup$ No; if they were independent, you would just multiply the probabilities. Clearly that is not the right thing to do here. You can literally read the answer off of the table given to you. $\endgroup$ – Aditya Dua Feb 12 at 1:03

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