A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 6?  Options: A) 3/8 B) 1/16 C) 1/8 D) 1/2 E) 0
One approach to solve the problem is to list the different possibilities for a toss of coin three times. Because there are two outcomes and the coin is tossed three times, the table will have 2*2*2 or 8 rows.   Next add the resulting rows together to find the sum (the fourth column in the table below).   Toss 1 | Toss 2 | Toss 3 | Sum   1 ---------- 1 -------- 1 ------ 3   1 ---------- 1 -------- 2 ------ 4   1 ---------- 2 -------- 1 ------ 4   1 ---------- 2 -------- 2 ------ 5   2 ---------- 1 -------- 1 ------ 4   2 ---------- 1 -------- 2 ------ 5   2 ---------- 2 -------- 1 ------ 5   2 ---------- 2 -------- 2 ------ 6   From the table we see that there are 0 situations where the sum of the tosses will be greater than 6. And there are 8 possible combinations resulting in a probability of   0/8 or a probability of 0.   SO the correct answer is 0.
<result>E</result>
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 6?