Question: A hat contains a number of cards, with 30% white on both sides, 50% black on one side and white on the other, 20% black on both sides. The cards are mixed up, then a single card is drawn at random and placed on the table. If the top side is black, what is the chance that the other side is white?
My attempt: Using the definition of conditional probability, we get, $P(W|B)=\frac{P(W\cap B)}{P(B)}=0.5/0.7=5/7$
But the actual answer is 5/9. Where did I go wrong?
On
One way to think about it is analyzing instead the events as which card is picked + its orientation. So instead of having 3 events (WW, BW, BB), we have 4 events (WW, BW, WB, BB).
Then the probabilities are
WW: 30%
BW: 25%
WB: 25%
BB: 20%
When we take a random card and place it in a random orientation, we see that it's black on top, so it can either be BW or BB. There is a 25% chance of BW happening and a 20% chance of BB happening, so we have $\frac{25}{25+20} = \frac{5}{9}$.
On
I'd think of it not as picking a card but picking a side (of some card). And let's pretend we have $100$ cards. So you randomly pick one of the $200$ sides. Then there are $50\cdot1+20\cdot2=90$ black sides. Of those, $50$ come with white on the other side. So $\frac{50}{90} = \frac{5}{9}$ chance for white other side.
It's not quite as simple.
The probability that the top side of the card chosen is black is not $0.7$ and here is why:
There is a $0.2$ chance that we pick a card that is black on both sides. Easy enough... However, when we bring in the $0.5$ chance of a black/white card, we need to factor in which side is faced upwards. It would seem natural to assume that either side could be up with equal probability. So there's a $0.5^2$ chance that a black/white card is chosen and the black side is faced up, bringing the total probability that black is the top side of the chosen card to $0.2 + 0.5^2$
Now that this fact has been brought to light perhaps you are able to finish the problem from here!