Question 1162362: 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 either the 2 of hearts, 2 of spades or 2 of diamonds; the second card played is the 3 of the same suit as the first card played, and the third card played is neither a club nor a heart.
Would the answer be: (3/51)*(1/50)*(26/49) = 0.00062425 ?
Thanks
Answer by greenestamps(13203) (Show Source):
You can put this solution on YOUR website!
Almost -- but not quite.
Let's look at the three factors in your calculation....
The first card is either the 2 of hearts, the 2 of spades, or the 2 of diamonds; and we now the fourth card is the 4 of clubs.
There are 51 possibilities for the first card; 3 of them are "good": 3/51, correct.
The second card is the 3 of the same suit as the first card played.
There are now 50 possible cards, and only one of them is the 3 of the same suit as the first card: 1/50, correct.
The third card is neither a club nor a heart -- i.e., it is either a spade or a diamond.
The 26/49 you show indicates that of the 49 cards remaining 26 are either a spade or a diamond. That's not quite right.
If the first card was the 2 of hearts, then there are indeed 26 cards left that are spades or diamonds. But if the first card was either the 2 of spades or the 2 of diamonds, then there are only 25 cards left that are spades or diamonds.
One way to account for this in calculating the desired probability is to make separate cases for the three different 2's on the first card.
(1) 2 of hearts, 3 of hearts, diamond or spade: (1/51)(1/50)(26/49)
(2) 2 of spades, 3 of spades, diamond or spade: (1/51)(1/50)(25/49)
(3) 2 of diamonds, 3 of diamonds, diamond or spade: (1/51)(1/50)(25/49)
Then the probability of the desired outcome is the sum of those three probabilities:

Another way to calculate the desired probability would be to say that the probability of getting a third card that is a diamond or a spade is

Then the calculation of the probability would be

Both calculations will lead to the same answer.
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