Respuesta :
Since the sum of the numbers on the three draws is 12, if we want the card numbered 2 to be drawn exactly two times, the third card can only be numbered 8. In fact, [tex] 2+2+8 = 12 [/tex], and there are no other possibilities, unless you consider the various permutations of the terms.
So, we have three favourable cases: we can draw 2,2,8, or 2,8,2, or 8,2,2. This are the only three cases where the card numbered 2 is drawn exactly two times, and the sum of the number on the three draws is 12.
Now, the question is: we have three favourable cases over how many? Well, we have 5 possible outcomes with each draws, and the three draws are identical, because we replace the card we draw every time.
So, we have 5 possible outcomes for the first draw, 5 for the second and 5 for the third. This leads to a total of [tex] 5 \times 5 \times 5 = 5^3 = 125 [/tex] possible triplets.
Once we know the "good" cases and the total number of possible cases, the probability is simply computed as
[tex] P = \cfrac{\text{number of favourable cases}}{\text{number of all possible cases}} = \cfrac{3}{125} [/tex]
