Probability of geting acceptance letter

bayes-theoremconditional probabilityprobability

Here is the problem I am currently working on:

"A high school student is anxiously waiting to receive letter telling her whether she has been accepted to a certain college. She estimates that the conditional probabilities of receiving notification on each day of next week, given that she is accepted and that she is rejected, are as follows:

  • P(Monday|accepted) = .15; P(Monday|rejected) = .05
  • P(Tuesday|accepted) = .20; P(Tuesday|rejected) = .10
  • P(Wednesday|accepted) = .25; P(Wednesday|rejected) = .10
  • P(Thursday|accepted) = .15; P(Thursday|rejected) = .15
  • P(Friday|accepted) = .10; P(Friday|rejected) = .20
    She estimates that her probability of being accepted is .6.
    "

There are currently two subproblems I am struggling with. The first one is following:

"If there is no letter through Wednesday, what is the conditional probability that she will be accepted?"

I interpreted this as "find $P(Acc|\bar W)$" (meaning "accepted supposing no letter came on Wednesday"). My logic is following:
$$P(Acc|\bar W) = \frac{P(Acc \cap \bar W)}{P(\bar W)} = \frac{P(Acc \cap \bar W)}{1 – P(Acc \cap W) – P(Rej \cap W)} = \frac{P(\bar W | Acc ) P(Acc)}{1 – P(Acc \cap W) – P(Rej \cap W)} = \frac{P(\bar W | Acc ) \cdot 0.6}{1 – 0.25 \cdot 0.6 – 0.1 \cdot 0.4}$$
My problem is that I don't know how to find $P(\bar W | Acc )$ in this equation. I had the idea of equating $P(\bar W \cap Acc ) = P(Acc) – P(Acc \cap W)$, but this doesn't give correct answer. Eventually, I gave up and looked for the answer to the problem, which stated that $P(\bar W \cap Acc ) = (0.6)^2$. Since $P(Acc) = 0.6$, I've decided that means that $P(\bar W | Acc) = 0.6$, meaning that if letter didn't came on Wednesday, we had no way of assigning probability except to revert to general $P(Acc) = 0.6$. With that in mind, I went for the following task:

"What is the conditional probability that she will be accepted if no letter arrives that week?"

Which I interpreted as "find $P(Acc|\bar F)$" (because if letter didn't came on Friday, it's safe to assume it didn't came in all previous days and didn't arrive this week), and applied the same logic: $$P(Acc|\bar F) = … = \frac{0.6^2}{1-0.1 \cdot 0.6 – 0.2 \cdot 0.4 } = \frac {6} {17} \neq \frac {9}{25} $$
which is correct answer. Because of this, it must be true that my initial assumption about that $P(\bar W | Acc) = 0.6$ meant was false. What I am missing to solve this problem?

(To note: correct answers to first and second problem are $\frac {12}{27}$ and $\frac {9}{25}$, respectively)

Best Answer

  • The initial probability of her being accepted but not on Monday, Tuesday or Wednesday is $0.6(1- 0.15-0.2-0.25) = 0.24$
  • The initial probability of her being rejected but not on Monday, Tuesday or Wednesday is $0.4(1-0.05-0.1-0.1) = 0.3$

So given no letter by Wednesday, her conditional probability of acceptance is $\frac{0.24}{0.24+0.3}=\frac{12}{27}$

Similarly

  • The initial probability of her being accepted but not on Monday, Tuesday, Wednesday, Thursday or Friday is $0.6(1- 0.15-0.2-0.25-0.15-0.1) = 0.09$
  • The initial probability of her being rejected but not on Monday, Tuesday, Wednesday, Thursday or Friday is $0.4(1-0.05-0.1-0.1-0.15-0.2) = 0.16$

So given no letter by Friday, her conditional probability of acceptance is $\frac{0.09}{0.09+0.16}=\frac{9}{25}$

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