Question 1117200
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The probability of getting "r" tails when flipping "n" coins is<br>
{{{C(n,r)/(2^n)}}}<br>
At most 1 tails means either 0 tails or 1 tails.  So we have<br>
{{{C(n,0)/(2^n)+C(n,1)/(2^n) = 3/16}}}
{{{1/(2^n)+n/(2^n) = 3/16}}}
{{{(n+1)/(2^n) = 3/16}}}
{{{16*(n+1) = 3*(2^n)}}}<br>
That equation can't be solved algebraically; but simple enumeration finds that n is 5:<br>
{{{16*(n+1) = 16*6 = 96}}}
{{{3*(2^n) = 3*(2^5) = 3*32 = 96}}}