Question 1149440
<br>
Since each card that is drawn is replaced before drawing the next card, the probability of each number on each draw is 4/12, or 1/3.  Then<br>
P(4) = P(2 and 2) = (1/3)(1/3) = 1/9
P(5) = P(2 and 3 OR 3 and 2) = (1/3)(1/3)+(1/3)(1/3) = 2/9
P(6) = P(2 and 4 OR 3 and 3 OR 4 and 2) = (1/3)(1/3)+(1/3)(1/3)+(1/3)(1/3) = 3/9
P(7) = P(3 and 4 OR 4 and 3) = (1/3)(1/3)+(1/3)(1/3) = 2/9
P(8) = P(4 and 4) = (1/3)(1/3) = 1/9<br>
Note that the sum of those probabilities is 9/9=1, giving us some reassurance that our method of calculating them is correct.<br>