You've doubled it.
Hello
You are dealt 4 cards (each card is numbered from 0 - 9). Each card is independed of the other card. What is the probability of getting 2 pairs?
P(2 pair) = 4C2P(1st = 2nd)P(3rd ≠ 2nd = 1st) P(4th = 3rd ≠ 2nd = 1st) = (6)(0.1)(0.9)(0.1)=0.054
Is this correct?
Thank you for your help. Just one more thing, did I make a mistake in the following calculations:
P(4 different digits) = P(2nd ≠1st)P(3rd ≠1st ≠2nd) P(4th ≠ 2nd ≠ 3rd ≠ 1st) = (0.9)(0.8)(0.7) = 0.504
P(pair) = (4 Choose 2)P(1st = 2nd)P(3rd ≠ 2nd = 1st)P(4th ≠ 3rd ≠ 2nd = 1st) = (6)(0.1)(0.9)(0.8) = 0.432
P(Three-of-a-kind) =(4 Choose 3)P(1st = 2nd = 3rd ≠ 4th) = (4)(0.1)2(0.9) = 0.036
P(Four-of-a-kind) = P(1st = 2nd =3rd =4th ) = 0.001