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?

Answer 1

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.