A hand of 13 cards is dealt from a standard deck of 52 cards. What is the
probability that it contains more aces than tens?
There are 4 aces, 4 tens, and 44 cards which are neither aces nor tens.
I've done the calculation for two of the cases. Do the others, then add them:
ACES TENS NEITHER PROBABILITY
1 0 12 (4C1)(4C0)(44C12)/(52C13) = 0.1328518567
2 0 11 (4C2)(4C0)(44C11)/(52C13) = ____________
2 1 10 (4C2)(4C1)(44C10)/(52C13) = ____________
3 0 10 (4C3)(4C0)(44C10)/(52C13) = ____________
3 1 9 (4C3)(4C1)(44C9)/(52C13) = ____________
3 2 8 (4C3)(4C2)(44C8)/(52C13) = 0.0066984129
4 0 9 (4C4)(4C0)(44C9)/(52C13) = ____________
4 1 8 (4C4)(4C1)(44C8)/(52C13) = ____________
4 2 7 (4C4)(4C2)(44C7)/(52C13) = ____________
4 3 6 (4C4)(4C3)(44C6)/(52C13) = ____________
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Total probability = ____________
How does this probability change when you have the information that the hand
contains at least one ace?It doesn't change at all, because that's automatically given. If there are
more aces than tens, there has to be at least one ace.
Edwin
