Question 1162362
<br>
Almost -- but not quite.<br>
Let's look at the three factors in your calculation....<br>
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.<br>
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.<br>
The third card is neither  a club nor a heart -- i.e., it is either a spade or a diamond.<br>
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.<br>
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.<br>
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.<br>
(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)<br>
Then the probability of the desired outcome is the sum of those three probabilities:<br>
{{{(1/51)(1/50)(26/49) + (1/51)(1/50)(25/49) + (1/51)(1/50)(25/49)}}}<br>
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<br>
{{{((1/3)(26/49) + (2/3)(25/49))}}}<br>
Then the calculation of the probability would be<br>
{{{(3/51)(1/50)((1/3)(26/49) + (2/3)(25/49))}}}<br>
Both calculations will lead to the same answer.<br>