How do I find the odds of drawing at least 1 Ace from 3 random picks of a deck of cards? I know this is short but I am really struggling with it. I have created a tree diagram like thing but once I multiply it all out it gives me the fraction 338/3849. This when multiplied by 13 (the amount of different numbers/faces in a deck of card) (Ace,1,2,3,4,5,6,7,8,9,10,Jack,Queen,King). Thanks for the help.

Guest Oct 23, 2017

#1**+2 **

With "at least" questions it is often worth thinking about "none" first!

The probability of drawing no Aces from 3 random picks: p_{none} = (48/52)*(47/51)*(46/50) →4324/5525

The probability of at least one Ace: p_{at least one} = 1 - p_{none} → 1 - 4324/5525 → 1201/5525 ≈ 0.217

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Alan Oct 23, 2017

#1**+2 **

Best Answer

With "at least" questions it is often worth thinking about "none" first!

The probability of drawing no Aces from 3 random picks: p_{none} = (48/52)*(47/51)*(46/50) →4324/5525

The probability of at least one Ace: p_{at least one} = 1 - p_{none} → 1 - 4324/5525 → 1201/5525 ≈ 0.217

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Alan Oct 23, 2017