Question

From a deck of five cards numbered 2, 4, 6, 8, and 10, respectively, a card is drawn at random and replaced. this is done three times. what is the probability that the card numbered 2 was drawn exactly two times, given that the sum of the numbers on the three draws is 12?

Answer 1

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, $2+2+8 = 12$, 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 $5 \times 5 \times 5 = 5^3 = 125$ possible triplets.

Once we know the "good" cases and the total number of possible cases, the probability is simply computed as

$P = \cfrac{\text{number of favourable cases}}{\text{number of all possible cases}} = \cfrac{3}{125}$