I have a bag containing 4 cards, and each card is either black or white, but you do not know the exact composition of the bag. Assume that from your point of view, it is initially equally likely that the bag contains 0, 1, 2, 3, or 4 white cards. I let you draw a single card from the bag at random and you find that the card is white. Thus, the question is: how many of the cards in the bag are white on the basis of this single observation?
- Compute P(2W|E), the probability there was exactly two cards in the bag, given the evidence E that a white card was drawn at random.
- Compute P(3W|E).
- Compute P(4W|E), the probability all four cards in the bag are white, given the evidence E that a white card was drawn at random.
- If you had to guess how many of the cards in the bag are white based on the observation of a single white card, what would be your best guess? (enter an integer between 0 and 4)
For the first problem I got 1/2 but that was incorrect
I think that there would be 5 possibilities despite the fact that there are 4 cards in the bag because all the cards could've been black despite the fact that a white card is pulled from the bag, but I might be overthinking stuff at this point.
This is reverse probability theorem . Let $E_i$ be the event that there are. $i$ cards white in bag for all $i=0,1,2,3,4$ since all events are equally likely then. $P(E_i)=\frac{1}{5}\,\,,\forall\,i=0,1,2,3,4$ $$. $$ Let. $A$ be the events that card drawn is white. $$\therefore P(A)=\sum_{I=0}^{4} P(E_i)P(A \vert E_i)$$ $$=0 +\frac{1}{5}\frac{1}{4}+\frac{1}{5}\frac{2}{4}+\frac{1}{5}\frac{3}{4}+\frac{1}{5}\frac{4}{4}$$ $$P(A)=\frac{1}{2}$$ And for 2 white card in bag known that card drawn is white is the probability will be $$P(E_2 \vert A)=\cfrac{P(E_2).P(A \vert E_2)}{P(A)}$$ $$=\cfrac{\frac{1}{5}\frac{2}{4}}{\frac{1}{2}}$$ $$=\frac{1}{5}$$