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A single series of flipping a coin lasts until the first head appears (let $X_i$ represent the waiting time for the first head in $i$-th series). Use CLT to estimate the smallest number of such series to ascertain with probability $0.9$ that the total amount of tails obtained in all flips exceeds $100$.

I am pretty sure that we have sort of binomial distribution here... but I don't know how to solve this. Thanks in advance.

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    What does CLT mean? – 2017-01-22
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    @Paul Central limit theorem – 2017-01-22

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Not the binomial distribution (since the number of trials in an individual series isn't fixed), but a close relative.

Supposing that $Y_i$ represents the number of tails that appear in the $i^\text{th}$ series, first try finding (or rather, approximating) $$ P\left(\sum_{i=1}^n Y_i \geq 100\right) $$ for fixed $n$. Then find the smallest value of $n$ such that the probability becomes at least 0.9.

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    $Y_n$ - number of tails that appear in the $n$-th series $P\left(Y_n=k\right)=P\left(X_n=k\right)=\frac{1}{2}\left(1-\frac{1}{2}\right)^k$ $EY_n=\sum_{k=0}^\infty k\cdot\frac{1}{2^{k+1}}=1 \\ \\ \text{Var}Y_n=\sum_{k=0}^\infty k^2\cdot\frac{1}{2^{k+1}}-1^2=2=\sigma^2 \\$ $P\left(\sum_{n=1}^nY_n>100\right)\ge0.9 \\ P\left(\frac{\sum_{n=1}^nY_n-nEY_1}{\sqrt{n}\sigma}>\frac{100-nEY_1}{\sqrt n \sigma}\right)\approx 1-\Phi\left(\frac{100-n}{\sqrt{2n}}\right)\ge0.9\\ \Phi\left(\frac{n-100}{\sqrt{2n}}\right) \le\Phi(1.29) \\ (n-100)^2=2n\cdot1.29^2\\ n_1\approx 88, \ n_2\approx114 $ – 2017-01-23
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    Is this solution correct? – 2017-01-23
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    Looks good. All that's left to do is to drop the extraneous root. – 2017-01-25