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This is a stupid question to ask but I need help with likeliness or probability. I don't really know what it is called in English. Basically the likeliness to hit a #6 on a dice is 1/6.

So lets say I have 4 Aces(cards) of each color on the table. What is the probability that:

Card #1 = Heart Card #2 = Diamond Card #3 = Club Card #4 = Spade 

Card #1 must be 1/4, #2 1/3, #4 1/2. And then what? I'm stuck :(

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You are on the right track: choice #3 has probability 1/2 to be the right one and once the three first choices are correct, the fourth one is automaically correct. So the probability you are after is 1/4 times 1/3 times 1/2, which is 1/24.

You can also consider that the whole order of the four cards is drawn at random amongst the 4! possible ones. By symmetry every possible order has the same probability hence the probability to get the one you want is 1/4!=1/24.

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    @user10426: The chance that two events happen is the product of the probability that the first happens, times the chance that the second happens assuming the first already did. For this, you might just try it. 1/4 of the time the first card is a heart. Assuming that it is, there are now three to choose from, so the chance that the second is a diamond is 1/3 and the chance the first two work is 1/12.2011-05-03