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A deck of Spanish cards has 48 cards, divided into four pints (clubs, cups, golds, and swords). From this deck, two cards are drawn simultaneously. Calculate the probability that?
a) Both are cups. b) At least one is a cup. c) One is a cup and the other is a sword.
Noelia Hermosilla
a) Both are cups.
The first draw has a 12/48 chance of being a cup. The second one has 11/47 (because we have one less card and one less cup) so 12/48 * 11/47 = 5.8%
b) At least one is a cup.
As with before, the first draw has a 12/48 chance of being a cup. But here's the thing. There's actually four possibilities. If C is CUP and N is NOT CUP then the four possible draws are: CC, CN, NC, and NN. Only in one do we have no cups. So what we want are the total odds of CC, CN, and NC.
CC = 12/48, then 11/47 = 5.8% (just like before)
CN = 12/48, then 36/47 = 19.1%
NC = 36/48, then 12/47 = 19.1%
Any of these things happening is a valid condition of at least one cup, so the total odds is any of them happening, 5.8+19.1+19.1 = 44%. Pretty good odds!
We can even verify by calculating the odds of NN:
36/48, then 35/47 = 55.8%
So my rounding has introduced some inaccuracy here. You'll need to run the numbers yourself if you need it to more decimal points, but the odds otherwise total correctly.
c) One is a cup and the other is a sword.
We basically just did this, since this is CN or NC. 19.1 + 19.1 gets us 38.2% chance that one or the other happens.
Michael Scalise
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