Question 1162347: Three cards are drawn without replacement from an ordinary deck of 52 playing cards. A fourth card is flipped, and known: the 6 of clubs.
The three cards are played in order. What is the probability that the first card played is the 2 of hearts, the second card played is the 3 of hearts, and the third card played is neither a club nor a heart.
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website! Three cards are drawn without replacement from an ordinary deck of 52
playing cards. A fourth card is flipped, and known: the 6 of clubs.
We know that the 6 of clubs was not drawn until the 4th. So it is impossible
that it could have been drawn before the 4th. So let's think of it not being
in the deck at the beginning, and there were only 51 drawable cards.
So in order to have been successful, knowing what we are given,
1. the 2 of hearts was drawn from 51 cards, with a probability of 1/51.
2. the 3 of hearts was then drawn from the remaining 50 cards, with a probability of 1/50.
3. the 3rd card was a spade or a diamond, and there were 26 of these among
the 49 remaining cards, (since no spades or diamonds were drawn in
successful cases), so the probability was 26/49.
Answer (1/51)(1/50)(26/49) = 26/124950 = 13/62475 = 2.080832333 × 10-4 =
0.0002080832333
Edwin
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