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If you were drawing cards w/o replacemnt from"Iraq's Most Wanted" deck what are the following probabilities: 1. U drw 2 cards (1 Saddam Hussein & 1 Chemical Ali) 2. U drw 14 cards and not one of them is Saddam Hussein - have 2 show work

2007-04-21 16:13:10 · 2 answers · asked by Michael S 1 in Science & Mathematics Mathematics

2 answers

Iraq's 55 most wanted.

There are 55 cards.

1. To draw first saddamn or Ali, that's 2 out of 55.
To then draw the second one, that's 1 out of 54

Multiply 2x1, which is 2. Multiply 55x54 which is 2970.
That's 2/2970, or 0.067%

2. 14 cards that are not saddam, easy.
54 x 53 x 52 x5 1 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41, which is 2.8 x 10^23 over 55^14, giving you 0.1220, or 12.2%

2007-04-21 16:21:07 · answer #1 · answered by monomat99 3 · 0 0

Probability of drawing 2 desired cards out of 52 is
(2/52)*(1/51) = 2/2652

Probability of drawing 14 cards and not getting a desired card is the probability of getting the desired card subtracted from 1, so
1-(1/52)*(1/51)*(1/50)...(1/38) (you do the math ☺)

HTH

Doug

2007-04-21 16:25:23 · answer #2 · answered by doug_donaghue 7 · 0 0

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