two cards are drawn from a 52card deck without replacement. what is the probability that the 2nd card is a heart?
The easy way is to see that it's just one out of four suits or .
That's because no matter what is done with the cards before drawing a card, as
long as the results of what's done with the cards before drawing that second
card is not given, then for any one card to come up a heart is still one out of
four suits.
However some people find that hard to grasp. So we can do it the long way.
There are 52 cards, 13 of which are hearts and 39 of which are non-hearts.
Let's describe the entire event, which is:
(1st a non-heart AND 2nd card a heart) OR (1st card a heart AND 2nd card a heart)
We want:
P[(1st a non-heart AND 2nd card a heart) OR (1st card a heart AND 2nd card a heart)]
Remembering that AND implies "multiply" and OR implies "add", we see that the
above equals:
P(1st a non-heart)*P(2nd card a heart) + P(1st card a heart)*P(2nd card a heart)
=
So it comes out the same the long way as when we immediately write the answer
as one suit out of 4 from the start.
Edwin