8 coins are tossed simultaneously. What is the probability of getting at least 6 heads?
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We'll use the binomial distribution. We have n=8 and p=q=0.5 (assuming that the coin is unbiased)
We have to find the probability of getting at least 6 heads i.e. we have to find P(X>=6) (where X is the number of heads).
P(X>=6)=P(X=6)+P(X=7)+P(X=8)
where
P(X=6)= 8C6 * (0.5)^6 * (0.5)^2 = 7/64
P(X=7)= 8C7 * (0.5)^7 * (0.5)^1 = 1/32
P(X=8)= 8C8 * (0.5)^8 * (0.5)^0 = 1/256
So the required probability is 0.1445
We have to find the probability of getting at least 6 heads i.e. we have to find P(X>=6) (where X is the number of heads).
P(X>=6)=P(X=6)+P(X=7)+P(X=8)
where
P(X=6)= 8C6 * (0.5)^6 * (0.5)^2 = 7/64
P(X=7)= 8C7 * (0.5)^7 * (0.5)^1 = 1/32
P(X=8)= 8C8 * (0.5)^8 * (0.5)^0 = 1/256
So the required probability is 0.1445
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