Question 1022437: A coin is flipped eight times in succession. In how many ways can at least five heads occur?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
A coin is flipped eight times in succession. In how many ways can at least five heads occur?
Solution:
Assume that a fair coin is used, so P(H)=P(T)=0.5.
We can then apply the binomial coefficients for r=5,6,7,8 in
P(r)=C(n,r)/2^n
where C(n,r) is the binomial coefficient (combinations) of r objects out of n.
The above equation is a special case of the binomial distribution where the probabilities of success and failure are both 0.5.
For r=5 to 8 heads, we need to sum the coefficients:
P(5)+P(6)+P(7)+P(8)
=0.21875+0.10938+0.03125+0.00391
=0.3633
It is interesting to note that in this particular case where P(H)=P(T), the distribution is symmetrical, so that
P(0)=P(8),
P(1)=P(7),
P(2)=P(6),
P(3)=P(5).
Therefore, P(5)+P(6)+P(7)+P(8)=P(0)+P(1)+P(2)+P(3)
Since P(0)+P(1)+P(2)....+P(8)=1.0, we have
P(5)+P(6)+P(7)+P(8)
=(1-P(4))/2 ..... which is much simpler to calculate than summing
=(1-0.27344)/2
=0.3633 as before.
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