probablility help please !

[aammmy33]

what is the probability of getting at least 1 diamond in a 5-card hand dealt from a standard 52-card deck?

i know that there's 13 diamonds in a 52 card deck. i used the formula n(E)/n(S) = 13 P 5 / 52 P 5 but the answer didnt make any sense. (p = permutation)

please help me out. thank you

 

[wjm11]

what is the probability of getting at least 1 diamond in a 5-card hand dealt from a standard 52-card deck?

The answer would be the sum of five probabilities: the probabilities of getting 1, 2, 3, 4, and 5 diamonds.

We could calculate each of these and add them, but a simpler approach is to ask oneself, “What’s the probability of not getting any diamonds in one’s hand?”

The probability of getting at least one diamond = 1-P(no diamonds).

There are 39 cards that are not diamonds, so:

P(no diamonds) = (39/52)(38/51)(37/50)(36/49)(35/48)

Make sense?

 

[pka]

what is the probability of getting at least 1 diamond in a 5-card hand dealt from a standard 52-card deck?

You should think in terms of combinations here.
The probability of no diamonds is \(\displaystyle \frac{_{39}\mathcal{C}_5}{_{52}\mathcal{C}_5}.\)
So what is the probability of at least one?