I got a deck of 52 cars, including 13 spades, hearts, diamonds and clubs. I select a card, put it on the side, without looking. I flip over the 2nd card, and then the 3rd card, looking at them. Both are spades. What is the probability (given this information) that the initially selected card is also spades?
I am supposed to work with Bayes theorem. After calculating i got the result ~5,98% But i am unsure if this is correct.
The notation here seems fine to me, so i would use it. None of those values are given. I have to get them myself, somehow.
P(S1∣S2,S3)=P(S1,S2,S3)/P(S2,S3)
P(S1,S2,S3) = 0,129 (i think) For P(S2,S3) i would change it first to P(S2|S3)*P(S3), then using bayes to get P(S2|S3) which is equal to P(S3|S2)*P(S2)/P(S3).
And from here on (maybe i did some mistakes already) i am quite lost. How do i get those values? What are the values for:
- P(S2)
- p(S3)
- P(S3|S2)
and why? How do i get those?
I'm not sure why you need Bayes here, as the question is equivalent to the following: out of a deck with 50 cards (only 11 spades) one is drawn at random. What is the probability that it's spades? The answer - $\tfrac{11}{50}=0.22$.
$P(S_2)$ is the probability that the second card is spades without further knowledge. But without any information, all cards are symmetric (there is no reason for the second place in a deck to favor any particular card or shape) so $P(S_2)=\tfrac{1}{4}$. Same goes for $P(S_3)$.
In a same fashion as the first paragraph, $P(S_2\vert S_3)=\tfrac{12}{51}$.