Question 1197528
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The wording is a bit confusing, but it sounds like there are 5 fair coins, and there's an extra 6th special coin which has 2 heads.


Since that special coin has 2 heads, this means we are guaranteed at least one head in the 6 coins. 
We just need to find the probability of getting exactly 1 head in the five fair coins.


We'll use the binomial distribution. 
n = 5 coins
x = 1 head exactly is what we're after
p = 0.5 is the probability of getting heads
{{{B(x) = expr((n!)/(x!*(n-x)!))*(p^x)*(1-p)^(n-x)}}}


{{{B(x) = expr((5!)/(x!*(5-x)!))*(0.5^x)*(1-0.5)^(5-x)}}}


{{{B(1) = expr((5!)/(1!*(5-1)!))*(0.5^1)*(1-0.5)^(5-1)}}}


{{{B(1) = 5*(0.5^1)*(1-0.5)^(5-1)}}}


{{{B(1) = 0.15625}}} This value is exact


The 0.5^1 portion represents the probability of getting that 1 head
(1-0.5)^(5-1) represents the probability of getting 4 tails
The 5 out front is the number of ways to get 1 head, ie 
HTTTT
THTTT
TTHTT
TTTHT
TTTTH


Answer: <font color=red>0.15625</font>
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