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In a swamp there are two regular frogs and three princes frogs, the queen take out $k$ frogs with return. Let $R$ be the number of times that regular frog took out.

Evaluate the minimal number k such that the probabilty to take out regular froge at the most $50\%$ from the tooks out will be at least $0.99$ using Chebyshev's inequality

My attempt:

$R\sim B(n,p)\;\;\;,\;\;\;$ $R\sim B(k,\frac25)\;\;\;,\;\;\;$ $E[R]=\boxed{\frac{2k}{5}}\;\;\;\;,\;\;\;\;\;$ Var$[R]=\boxed{\frac{6k}{25}}$

$$P(|R-E[R]| \geq \alpha)\leq \frac{\text{Var}(R)}{\alpha ^2}$$

$$P\big(\big|R-\frac{2k}{5}\big| \geq \alpha\big)\leq \frac{6k}{25 \alpha ^2}$$

Is it correct so far? I don't know how to proceed, hints please

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Hints:

  • $\dfrac{k}{2} - \dfrac{2k}{5} = \dfrac{k}{10}$

  • $1-0.99 = 0.01$

so see what you should use for $\alpha$, and so what $k$ should be.

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  • $\begingroup$ Not relevant to finding the answer to the question as asked, but the resulting $k$ using Chebyshev's inequality is about $18$ to $20$ times as large as the result you could get from the binomial distribution. $\endgroup$ – Henry Aug 5 '15 at 11:30
  • $\begingroup$ $P\bigg(R-\frac{2k}{5}\geq \frac{k}{10}\bigg)\leq \frac{600}{25k}$ how can I proceed? $\endgroup$ – 3SAT Aug 5 '15 at 19:27
  • $\begingroup$ $0.01\leq \frac{600}{25k},\;\;\;\;k\leq 2400$ is it correct? $\endgroup$ – 3SAT Aug 5 '15 at 19:35
  • $\begingroup$ @Nehorai: Yes - I think $2400$ is the expected result except that the inequality is reversed so $k \ge 2400$. Earlier you wanted $ P\bigg(R-\frac{2k}{5} \geq \frac{k}{10}\bigg)\leq 0.01$ which is true if $\frac{600}{25k}\le 0.01$. $\endgroup$ – Henry Aug 5 '15 at 22:53
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    $\begingroup$ Using R and the binomial distribution which(pbinom((1:150)/2, size=1:150, prob=2/5) >= 0.99) the true cases are $k=122,124,126,128, 130$ or $k \ge 132$. This shows that $k\ge 2400$ is a very weak result, as often happens with Chebyshev's inequality. $\endgroup$ – Henry Aug 5 '15 at 22:57

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