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When run, a computer program chooses the word "expensive'' with probability 2/3 and otherwise chooses the word "free''. It then selects a letter uniformly at random from the word it chose and outputs that letter. Given the program outputs "e'', what is the probability that it chose the word "free''?

What I tried : Probability of the word "expensive" is 2/3, so the probability of getting the word "free" is 1/3. So from the word "free" there is 2 "e", so I guess the probability of getting the letter "e" is 1/2.

The answer that I got is 5/6, which I don't think so its correct. I need some help on this

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    $\begingroup$ This is a question of conditional probability. ''So from the word "free" there is 2 "e", so I guess the probability of getting the letter "e" is 1/2.'' but e is also present in expensive, thus you have to make cases $\endgroup$
    – user1012971
    Commented Jun 2, 2022 at 5:52
  • $\begingroup$ @RamanujanXXV Oh, I misread the question, Thanks for pointing it out ! $\endgroup$
    – Ben
    Commented Jun 2, 2022 at 5:54

2 Answers 2

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We can use Bayes Theorem to solve this question. Let $X$ be the random variable that outputs the word $(\text{expensive/free})$ and let $L$ be the random variable that outputs the letter. We are trying to calculate $P(X=\text{free}|L=\text{e})$. Using Bayes Theorem: $$ \begin{align} P(X=\text{free}|L=\text{e}) &= \dfrac{P(X=\text{free})P(L=\text{e}|X=\text{free})}{P(L=\text{e})}\\ &= \dfrac{\frac{1}{3}\times \frac{1}{2}}{\frac{2}{3}\times \frac{3}{9}+\frac{1}{3}\times \frac{1}{2}}\\ &= \dfrac{\frac{1}{6}}{\frac{7}{18}}\\ &=\dfrac{3}{7} \end{align} $$

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$$P(free|e)= \frac{P(free)\cap P(e)}{P(e)}$$ By the total probability theorem,$ P(e) = P(e|free).P(free) + P(e|expensive)P(expensive) = \frac{2}{3} \frac{1}{3} + \frac{1}{3}\frac{2}{4} = \frac{7}{18}$. Thus $$P(free|e)=\frac{\frac{1}{2}\frac{1}{3}}{\frac{7}{18}}= \frac{3}{7}$$.

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