+0 n coins are simultaneously flipped. The probability that at most one of them shows tails is 5/16. Find n. Thanks !
+399 The probability that at most one of them shows tails = the probability that none of them show tails (all of them are heads) + the probability that one of them shows tails.
For any n, the probability that none of them shows tails is \({(\frac{1}{2})}^{n}\), because heads would needed to be landed on every single time.
The probability that one of these show tails is more difficult. Suppose the first one is tails, and the rest are heads, but we also note that any of these can be the one that lands on heads, so there are n ways to arrage these, so overall the probability is \(n*{(\frac{1}{2})}^{n}\).
Adding these together - \(n*{(\frac{1}{2})}^{n}+{(\frac{1}{2})}^{n}=(n+1){(\frac{1}{2})}^{n}=\frac{5}{16}\), \((n*1)*(\frac{1}{{2}^{n}})=\frac{n+1}{{2}^{n}}=\frac{5}{16}\). We see one such value that satisfies our equality is n = 4. But we know the rate of increase for exponential functions is exponetially increasing/decreasing, so after that the gap between n+1 and 2n, will only constantly increase, therefore our only positive solution is 4.
