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This is just a sort of clarification question: suppose I wanted to know the probability of getting at least one ace in a 2card hand. My question is : can this be calculated directly or would I need to use the complement?

Calculating directly, I said for the first card of the two there are 4 possibilities and for the second there are 51 possibilities so the required amount of hands is 4x51. I think this needs to be divided by 2! because I am not considering order.

The probability would then be [tex] \frac{\frac{4×51}{2!}}{(52 choose 2)}, [/tex] could someone tell me whether this is correct or not. When I use it as part of a later question, I get the wrong answer.

Thanks

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# Homework Help: 2 card hand probability question

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