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?

  1. 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.
  2. Compute P(3W|E).
  3. 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.
  4. 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.

  • $\begingroup$ If all the cards in the bag were black then the probability that you drew a white one would be $0$. $\endgroup$ – lulu Feb 15 at 1:44
  • $\begingroup$ yes I know that...that was actually the one question I got right so I didn't include it. I just don't know if I should keep using the 1/5 probability anymore since it is impossible. $\endgroup$ – mad Feb 15 at 1:46
  • $\begingroup$ To visualize the problem, suppose you had $100$ bags, $20$ all black, $20$ with one white and three black, etc. You reach into a bag at random and draw a white card. Now you know it wasn't one of those first $20$. Could it have been in the second $20$? Sure, but it's not likely. Indeed, you'd expect to get a white card from only $5$ of those $20$. Now analyze the third group of $20$, and so on. $\endgroup$ – lulu Feb 15 at 1:49
  • $\begingroup$ You could, of course, just use Bayes' Theorem...but I think that visualization of Bayes' Theorem might be useful. $\endgroup$ – lulu Feb 15 at 1:49

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 $$P(E_2 \vert A)=\cfrac{P(E_2).P(A \vert E_2)}{P(A)}$$ $$=\cfrac{\frac{1}{5}\frac{2}{4}}{\frac{1}{2}}$$ $$=\frac{1}{5}$$

  • $\begingroup$ Please replace fractions by conditional probabilities $P(A \mid E_i)$. $\endgroup$ – NCh Feb 15 at 2:13
  • $\begingroup$ I mean conditional only but you are correct I have typed as fraction I will correct that. $\endgroup$ – Rajan Feb 15 at 2:16
  • 1
    $\begingroup$ Corrected it thanks $\endgroup$ – Rajan Feb 15 at 2:18

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