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We toss $n=200$ Euro coins on the table.

I want to calculate, using the central limit theorem, the probability that at least $110$ coins have tutned on the same side.

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Do we have to consider one side, without loss of generality, to determine the random variables $X_i$ as follows: \begin{equation*}X_i=\begin{cases} 1 , & \text{coin } i \text{ shows head}\\ 0 , & \text{coin } i \text{ does not show head} \end{cases}\end{equation*}

Or do we not have to consider one specific side of the coin? But how would the random variables be then defined?

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  • $\begingroup$ The probability that at least $110$ show $H$ is the same as the probability that at least $110$ show $T$ so if you just do one side all you have to do is double the result. $\endgroup$ – lulu Jan 15 '18 at 14:12
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You can define $X_i$ as you suggest though not all Euros have a head (is it the map side they all have? or the other side which sometimes has a head?) Let's define $X_i$ as an indicator of the map side, so you either want at least $110$ or no more than $90$ map sides showing

You then want $\mathbb P \left(\sum X_i \ge 110\right)+\mathbb P \left(\sum X_i \le 90\right)$

Assuming independence, you can then use the binomial distribution for an exact calculation, or a Gaussian approximation with a cuttoffs at $90.5$ and $109.5$ (or using symmetry, double no more than $90.5$). The probability would be almost $18\%$

For example in R:

> pbinom(90, size=200, prob=0.5) + 1 - pbinom(109, size=200, prob=0.5)
[1] 0.178964

> 2 * pnorm(90.5, mean=200*0.5, sd=sqrt(200*0.5*0.5)) 
[1] 0.1791092
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  • $\begingroup$ I understand! I wnat also to calculate which has to be at least the value of $n$ such that with probability $\geq 0.95$ at at least $45\%$ und at most at $55\%$ of the coins "head" is shown. I have to use again the CLT. Is it as follows? \begin{align*}\mathbb{P}\left (n\cdot 45\%\leq S_{n}\leq n\cdot 55\%\right )\geq 0.95 &\Rightarrow \mathbb{P}\left ( S_{n}\leq n\cdot 0.55\right )-\mathbb{P}\left ( S_{n}\leq n\cdot 0.45\right )\geq 0.95 \\ & \Rightarrow \Phi \left (\frac{ 0.55n-n\mu }{\sqrt{n\sigma^2}}\right )-\Phi \left (\frac{ 0.45n-n\mu }{\sqrt{n\sigma^2}}\right )\geq 0.95\end{align*} $\endgroup$ – Mary Star Jan 15 '18 at 15:20
  • $\begingroup$ If this is correct how could we continue? $\endgroup$ – Mary Star Jan 15 '18 at 15:21
  • $\begingroup$ @MaryStar Use $\Phi \left (\frac{ 0.45n-0.5n }{\sqrt{n\sigma^2}}\right )=1-\Phi \left (\frac{ 0.55n-0.5n }{\sqrt{n\sigma^2}}\right )$ and then the inverse standard normal cdf $\Phi^{-1}()$ $\endgroup$ – Henry Jan 15 '18 at 17:19
  • $\begingroup$ So, does it hold that $\Phi (a)-\Phi (b) = \Phi (b-a)$ Or how do get that? $\endgroup$ – Mary Star Jan 15 '18 at 17:25
  • $\begingroup$ @MaryStar That looks very wrong $\endgroup$ – Henry Jan 15 '18 at 17:27
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By using the CLT:

Let $\sum_{i=1}^{200} X_i$ be the number of "heads" in $200$ coin flips. Hence, $\sum X_i \sim Bin (n,p)$ or $$ \sum X_i\overset{approx.}{\sim}\mathcal{N} (np, np(1-p)), $$ thus \begin{align} \mathbb{P}( \sum X_i \ge 110) &= 1- \mathbb{P}( \sum X_i < 110)\\ &\approx 1- \Phi\left( \frac{109.5 - np }{\sqrt{npq}} \right). \end{align} For tails it will be the same calculations. Thus assuming a fair coin and independent tosses, the probability of the event if interest approx equals $$ 2 \left( \approx 1- \Phi\left( \frac{109.5 - 100 }{\sqrt{50}} \right) \right) $$

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  • $\begingroup$ Ah I see! I wnat also to calculate which has to be at least the value of $n$ such that with probability $\geq 0.95$ at at least $45\%$ und at most at $55\%$ of the coins "head" is shown. I have to use again the CLT. Is it as follows? \begin{align*}\mathbb{P}\left (n\cdot 45\%\leq S_{n}\leq n\cdot 55\%\right )\geq 0.95 &\Rightarrow \mathbb{P}\left ( S_{n}\leq n\cdot 0.55\right )-\mathbb{P}\left ( S_{n}\leq n\cdot 0.45\right )\geq 0.95 \\ & \Rightarrow \Phi \left (\frac{ 0.55n-n\mu }{\sqrt{n\sigma^2}}\right )-\Phi \left (\frac{ 0.45n-n\mu }{\sqrt{n\sigma^2}}\right )\geq 0.95\end{align*} $\endgroup$ – Mary Star Jan 15 '18 at 15:20
  • $\begingroup$ If this is correct how could we continue? $\endgroup$ – Mary Star Jan 15 '18 at 15:20
  • $\begingroup$ Use $ \Phi(-\alpha) = 1 - \Phi(\alpha)$ and then you will have to solve a quadratic equation w.r.t. $n$. $\endgroup$ – V. Vancak Jan 15 '18 at 15:27
  • $\begingroup$ I got it! Thanks! I have also an other question at an other problem: $$$$ A hotel has 800 beds and it is known that 10% of the bookings are canceled. To achieve a complete occupancy, the management decides to confirm more bookings than beds are available. We assume that people decide independently whether they make their booking. Using the CLT I want to calculate the probability that all people that show up in the hotel, for whom a bed has been reserved, will also receive a bed if 820 bookings have been confirmed. $\endgroup$ – Mary Star Jan 15 '18 at 20:17
  • $\begingroup$ I have defined the random variable $X_i$ with \begin{equation*}X_i=\begin{cases} 1 , & \text{person } i \text{ is showing up}\\ 0 , & \text{person } i \text{ is not showing up} \end{cases}\end{equation*} We have that $\mu=p=0,9$ and $\sigma^2=p(1-p)=0,9\cdot 0,1=0,09$. $\endgroup$ – Mary Star Jan 15 '18 at 20:17

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