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I have this exercise. it involves Bayes' Theorem:

in an exam, $29\%$ of students chose to write essay $A,$ if a student chooses to write essay $A,$ then student had read author $X,$ and $P(X \text{ given } A) = 0.90.$

if a student chooses to write a different essay, then student didn't read author $X,$ and $P(\text{not }X \text{ given not }A) = 0.90.$

if we select randomly a student who read author $X,$ exercise asks to find $P(A \text{ given } X).$ (i.e a student who wrote essay $A$ given $X$.)

I've used Bayes' theorem: $$P(A|X) = \frac{P(X|A) * P(A)}{P(X|A) * P(A) + P(X|\text{not }A) * P(\text{not }A)}$$

the problem is in this part: $P(X|\text{not }A)$, I don't know how to find it.

I know these data: $P(A) = 0.29$, $P(\text{not }A) = 0.71$, $P(X|A) = 0.90$, $P(\text{not }X|\text{not }A) = 0.90$.

Hypothesis: I think $P(X|\text{not }A)$ is equal to $0.10,$ because $P(\text{not }X|\text{not }A) = 0.90$. But I'm trying to formalize that thought.

Stephen Donovan
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Gabriel Burzacchini
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    Related: https://math.stackexchange.com/questions/2279851/applied-probability-bayes-theorem/2279888#2279888 – Ethan Bolker Jun 11 '22 at 12:07
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    Formally: Conditional probability measures obey all the rules for probability measures, including the Additive Rule for Complementary Events. $$\mathsf P(X\mid\text{not }A)+\mathsf P(\text{not }X\mid\text{not }A)=1$$ – Graham Kemp Jun 11 '22 at 12:20

3 Answers3

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Your intuition is correct, for any events $A$ and $B$ with $P(B) \neq 0,$ we have $P(A | B) + P(\text{not } A | B) = 1.$

We can show this rather easily with the definition and the law of total probability: noting that $A \cap B$ and $\text{not } A \cap B$ must be disjoint, we have

$$\begin{align}P(A|B) + P(\text{not } A | B) & = \frac{P(A \cap B)}{P(B)} + \frac{P(\text{not } A \cap B)}{P(B)} \\ & = \frac{P\big((A \cap B) \cup (\text{not } A \cap B)\big)}{P(B)}\\ & = \frac{P\big((A \cup \text{not } A) \cap B\big)}{P(B)} \\ & = \frac{P(B)}{P(B)} = 1\end{align}$$

(I've also written in the steps which justify the law of total probability for your convenience: if you're comfortable with the fact that $P(A \cap B) + P(\text{not } A \cap B) = P(B)$ then you can skip the steps in the middle lines.)

Stephen Donovan
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First find $P(X)$ by using: $P(\bar{X}|\bar{A})=\frac{P(\bar{X}\cap\bar{A})}{P(\bar{A})}=\frac{1-P(X\cup A)}{1-P(A)}=0.9$ where $P(X\cup A)=P(X)+P(A)-P(X\cap A)$ and since $P(X\cap A)=P(A)\cdot P(X| A)$ the only unknown is $P(X)$. Then use the fact that $P(A|X)=\frac{P(X|A)P(A)}{P(X)}$ by bayes and since we found $P(X)$ previously we are done.

Omer
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Imagine 1000 students. 29%, 290 students choose to write essay A, the other 1000- 290= 710 do not. 90% of the students who choose to write essay A, 261 read X, the other 290- 261= 29 do not.

Of the 710 students who did not choose essay A, 90%, 639 students did NOT read author X, 710- 639= 71 read author X.

So a total of 261+ 71= 332 students read author X of whom 261 chose essay A. The probability a student chose essay A given that he read author X is 261/332= 78.6%.

George Ivey
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