What is the probability of drawing a diamond or an ace from a deck of cards

I think it's 26/52 or 2/4 since it's either. 

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  well let's see. you have 52 cards in a deck. you have 4 different "symbols" or whatever (ace, diamond blah blah) . just divide it by 4. the quotient is 13. since you're looking for two symbols (diamond and ace), multiply it by two and you have 26 (you did say either) .

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  if i'm not mistaken, in probability, it' always in fraction form? so yeah. 2/4 is just changing it to it's simplest form

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  Ahhh!!!! Thank you!!! ^_^

• Probability is also in lowest term 26/52 = 1/2, or 50% probability.

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  its lowest term is 1/2 or 50% probability

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Ok I readjusted my solutions a bit.

Solution 1: We have two cases because if the first card is a $\diamondsuit$, it could be an ace or not be an ace.

There is a $\dfrac{1}{52}$ chance that the ace of $\diamondsuit$ is drawn first, and a $\dfrac{3}{51} = \dfrac{1}{17}$ chance that the second card drawn is one of the three remaining aces, which gives a probability of $\dfrac{1}{52}\cdot \dfrac{1}{17} = \dfrac{1}{884}$ chance that this occurs.

There is a $\dfrac{12}{52} = \dfrac{3}{13}$ chance that a $\diamondsuit$ other than the ace is drawn first, and a $\dfrac{4}{51}$ chance that an ace is drawn second, giving a $\dfrac{3}{13}\cdot \dfrac{4}{51} = \dfrac{4}{221}$ chance that this occurs.

So the probability that one of these two cases happens is $\dfrac{1}{884} + \dfrac{4}{221} = \boxed{\dfrac{1}{52}}$.

Notice that we can avoid some of the large denominators above by organizing this computation as follows:$$\dfrac{1}{52}\cdot\dfrac{3}{51}+\dfrac{12}{52}\cdot\dfrac{4}{51} = \dfrac{1\cdot 3+12\cdot 4}{52\cdot 51} = \dfrac{51}{52\cdot 51}=\boxed{\dfrac{1}{52}}.$$ Solution 2: We can solve this problem by using symmetry. Make a new "card" by combining the suit of the first card with the rank of the second card; for example, if the first two cards are $\spadesuit$2 and $\heartsuit$Q, then the new "card" would be $\spadesuit$Q. Then this new "card" is equally likely to be any of $52$ possibilities, one of which is the desired $\diamondsuit$A. So, the probability that the first card is a $\diamondsuit$ and the second card is an ace is $\boxed{\frac1{52}}$.

You can put this solution on YOUR website!

There are 13 diamonds in a deck. There are 4 aces in a deck. There is one ace of diamonds in a deck. The number of cards that are either diamonds or aces or both are,

The probability of drawing a diamond from a standard deck of 52 cards is 13 in 52, or 1 in 4, or 0.25.