0
$\begingroup$

Assume we have n number of sets with varying number of elements, some of which appear in more than one set. Each element can have zero or more attributes.

If we are told that an element appears in all the sets, what is the probability that it has a specific attribute t?

As a concrete example, let's say we have two decks of standard playing cards, one missing the diamonds and one missing the clubs as well as spades. Now if a card that appears in both decks is picked, what is the probability that it's a heart?

$\endgroup$
  • $\begingroup$ You ask for a conditional probability. Ergo, the only elements that need consideration are those appearing in all the sets. Accordingly, you don't have to worry about $n$ sets: you just have one set; namely, their intersection. At this point objective analysis of your question ends, because it makes no sense to provide a probability (except in trivial cases where all or none of the elements of the union have attribute $t$) until you describe how a given element in the union was selected. What can you tell us about that procedure? $\endgroup$ – whuber Sep 8 '13 at 16:21
1
$\begingroup$

In your concrete example, the probability is 1, since the only cards that appear in both decks are hearts.

For a harder example, suppose deck 1 has clubs and spades and deck 2 has diamonds, clubs, and spades. Then what is the probability that a card is a spade, given that it is in both decks? This now depends on which deck it was picked from. If it was picked from deck 1, then it can only be a club or a spade, so the probability is 1/2 that it is a spade. If from deck 2, the probability is 1/3

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Peter, it's already given that the card appears in both decks. My questions is, what is the overall probability that the card is of a given suit. It is quite trivial to figure out the probability for each individual deck. $\endgroup$ – Mansour Feb 10 '13 at 15:29
  • $\begingroup$ For Peter's example, if we have $P(Deck=1)=P(Deck=2)=\frac{1}{2}$, we have $$P(Card=Spade) = P(Card=Spade|Deck=1)P(Deck=1)+P(Card=Spade|Deck=2)P(Deck=2) = \frac{1}{2}\frac{1}{2}+\frac{1}{3}\frac{1}{2}=\frac{5}{12}=0.416$$. That is what you asked for? $\endgroup$ – rapaio Aug 9 '13 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.