We flip a fair coin 10 times. What is the probability that we get heads in at least 6 of the 10 flips?
+4622 Each coin has two sides, heads or tails, so the total number of possible outcomes is \(2^{10}=1024\).
The number of ways of getting heads is 10 C 6 * 10 C 7 + 10 C 8 + 10 C 9 + 10 C 10=210+120+45+10+1=386 ways.
193/512 but I might be worng
10! / [6!*4!*2^10] + 10! / [7!*3!*2^10] + 10! / [8!*2!*2^10] + 10! / [9!*1!*2^10] + 10! / [10!*2^10] =193 / 512 =37.6953125%
+129852 At least 6 heads means
6 heads = C(10,6) (,5)^10
7 heads = C(10, 7) (.5)^10
8 heads = C(10,8)(.5)^10
9 heads = C(10,9) (,5)^10
10 heads = (.5)^10
So....the total probability is
(.5)^10 [ C(10,6) +C(10,7) + C(10,8) + C(10,9) + 1 ] ≈ .3769 ≈ 37.69%
+1252 6H, 4T = 10!/6!4! ways to arrange
7H, 3T = 10!/7!3!
8H, 2T = 10!/8!/2!
9H, 1T = 10!/9!1!
10H, 0T = 10!/10!0!
Add it up to get 386
2^10 possible outcomes = 1024
Answer: 193/512
You are very welcome!
:P
