Answer: B) 4%
=====================================================
Explanation:
We'll involve the nCr combination formula here. This is because order doesn't matter.
There are n = 13 diamonds and we want to select exactly r = 3 of them.
So,
[tex]_n C _r = \frac{n!}{r!*(n-r)!}\\\\_{13} C _{3} = \frac{13!}{3!*(13-3)!}\\\\_{13} C _{3} = \frac{13!}{3!*10!}\\\\_{13} C _{3} = \frac{13*12*11*10!}{3!*10!}\\\\_{13} C _{3} = \frac{13*12*11}{3!}\\\\_{13} C _{3} = \frac{13*12*11}{3*2*1}\\\\_{13} C _{3} = \frac{1716}{6}\\\\_{13} C _{3} = 286\\\\[/tex]
There are 286 ways to pick the three diamond cards. Then there are 52-13 = 39 ways to pick the fourth card that is either a spade, club of heart.
So we have 286*39 = 11,154 ways to pick the four cards given the conditions your teacher set.
This is out of 52C4 = 270,725 ways to pick four cards (use the nCr formula above with n = 52 and r = 4).
Divide the two values to get the answer:
(11,154)/(270,725) = 0.04120048019208
that rounds to 0.04 which converts to 4%
So there's roughly a 4% chance of getting exactly 3 diamond cards.
