How many different sequences of ten flips of a coin have at least 8 heads?
We find the number of sequences with 8 heads, 9 heads, and 10 heads
separately and add these totals. There are ${10\choose 8} = {10\choose 2} =
45$ sequences with 8 heads, ${10\choose 9} = {10\choose 1} = 10$ sequences
with 9 heads, and only 1 with 10 heads. Thus, there's a total of
$45+10+1 = 56$ sequences of ten flips with at least 8 heads
To calculate the number of different sequences of ten flips of a coin that have at least 8 heads, we can break down the problem into cases.
Case 1: Exactly 8 heads
In this case, we have 8 heads and 2 tails. The number of ways to arrange these outcomes can be calculated using combinations. We need to choose 8 positions out of the 10 flips to be heads, so the number of ways is given by the formula C(10, 8) = 45.
Case 2: Exactly 9 heads
Similarly, we have 9 heads and 1 tail. The number of ways to arrange these outcomes is given by the formula C(10, 9) = 10.
Case 3: All 10 flips are heads
In this case, we have all 10 flips as heads. There is only one way to arrange this outcome.
Therefore, the total number of different sequences of ten flips with at least 8 heads is the sum of the results from each case: 45 + 10 + 1 = 56.
So, there are 56 different sequences of ten flips of a coin that have at least 8 heads.