2007 HSC Q9 - Probability Question (1 Viewer)

[Lucas_]

A pack of 52 cards consists of 13 cards of each suit.

ii) Four cards are chosen at random from the pack and placed side by side and the table. What is the probability that these four cards are all from different suits?

--> 52/52 x 39/51 x 26/50 x 13/49 = 0.105 (to 3 d.p) is the correct answer,
can somebody please elaborate on why this does not need to be multiplied by 4 (as it could happen with any of the 4 suits?)

Thanks as always

 

[asianese]

By choosing one card at a time, you eliminate the other suits while you do so. So you don't need to multiply by 4.

So e.g. if you choose a heart, you choose it so there is 52/52. Then you need to choose any other suit from 51, so 39/51 and then there are only two suits left now. 26/50 remain to be chosen, and the last suit has 13/49.

 

[Lucas_]

Thanks!