$n$ coins are simultaneously flipped. The probability that at most one of them shows tails is $\frac{3}{16}$. Find $n$.

We have that either none of them show tails or one does..so....

P(none show tails) =   C(n, 0) (1/2)^n =  (1/2)^n

P( 1 shows a tail) = C(n, 1) (1/2)^n  = n* (1/2)^n

So we have that

(1/2)^n ( n + 1) = 3/16

(n + 1) / 2^n =   3 /  2^4

2^4 / 2^n   =      3 / (n + 1)

2^n / 2^4  =   (n + 1) / 3

2^( n - 4)   =  (n + 1) / 3

Note that the left side is some power of 2

So.....the first positive integer n that makes the right side a power of 2 is when n = 2

But this gives   2^(-2)  not equal to  3/3

The next positive integer that makes the right side a power of 2 is when n  = 5 

So

2^(5 - 4)  =  (5 + 1) /3

2 = 6/3     is true

So.....we need 5 coins