# Probability to get a chocolate snowman or a chocolate reindeer

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

$N\in \mathbb{N}$ christmas presents will be distributed. In every gift there is additional either a chocolate snowman or a chocolate reindeer. It will be independent of each other and with the same probability of giving out gifts with a chocolate snowman or chocolate reindeer. So let $X_1, \ldots , X_n$ be independent and Bernoulli-distributed with probability of success $p =\frac{1}{2}$. The event $\{X_j = 1\}$ means that in the $j$th present there is a chocolate reindeer. Let $a_n$ be the probability that at least 60% of the $n$ gifts distributed contains a chocolate reindeer.
(a) Calculate $a_{10}$ explicitly. Enter intermediate steps.
(b) Use the inequality $\displaystyle{P\left [\left |\frac{1}{n}\sum_{j=1}^nX_j-p\right |\geq \epsilon\right ]\leq \frac{1}{4n\epsilon^2}}$ from Bernoulli's weak law of large numbers, to estimate $a_{100}$. Does it hold $a_{100}<a_{10}$ ?
(c) Determine $\displaystyle{\lim_{n\rightarrow \infty}a_n}$.

For (a) we have that \begin{equation*}P(a_{10}\geq 0.6)\Leftrightarrow P\left(\sum\limits_{i=1}^{10} X_i\geq 0.6\cdot 10 \right)\Leftrightarrow P\left(\frac{1}{10}\cdot \sum\limits_{i=1}^{10} X_i\geq 0.6\right)\end{equation*}
Do we have to use the weak law of large numbers $P[|\overline{X}-\mu|\geq \epsilon]\leq \frac{\sigma^2}{n\epsilon^2}$ ?
The expected value of $X_i$ is $\mu=E[X_i]=p=\frac{1}{2}$ and the variance is $\sigma^2=\text{Var}(X_i)=p\cdot (1-p)=\frac{1}{4}$. So we get $$P[|\overline{X}-\mu|\geq \epsilon]\leq \frac{\sigma^2}{n\epsilon^2} \Rightarrow P\left [\left |\overline{X}-\frac{1}{2}\right |\geq \epsilon\right ]\leq \frac{\frac{1}{4}}{10\cdot \epsilon^2}\Rightarrow P\left [\overline{X}\geq \epsilon+\frac{1}{2}\right ]\leq \frac{\frac{1}{4}}{10\cdot \epsilon^2}$$ For $\epsilon=0.1$ we get $$P\left [\overline{X}\geq 0.6\right ]\leq \frac{\frac{1}{4}}{10\cdot 0.1^2} \Rightarrow P\left [\overline{X}\geq 0.6\right ]\leq 2.5$$
This cannot be correct, can it?

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#### GJA

##### Well-known member
MHB Math Scholar
Hi mathmari ,

Great job rewriting the condition as $a_{10} = \text{Pr}\left(\displaystyle\sum_{i=1}^{10}X_{i}\geq 6\right).$ Your attempt to use Bernoulli's weak law of large numbers is also good. Note, though, that Bernoulli only gives an upper bound for $a_{n}$, so it will not allow us to calculate $a_{10}$ explicitly.

To proceed, expand your previous equation one step further to $$a_{10} = \text{Pr}\left(\sum_{i=1}^{10}X_{i}\geq 6\right) = \text{Pr}\left(\sum_{i=1}^{10}X_{i} = 6\right) + \text{Pr}\left(\sum_{i=1}^{10}X_{i} = 7\right) + \cdots + \text{Pr}\left(\sum_{i=1}^{10}X_{i} = 10\right).$$ To calculate each of the 5 terms on the right, consider using the Binomial Distribution, which calculates the probability of $k$ successes in a sequence of $n$ independent Bernoulli experiments.

Nice work so far. Feel free to let me know if you have any other questions.

Edit: Changed $\text{Pr}(a_{10}\geq0.6)$ to $a_{10}$ because $a_{n}$ is, by definition, the probability of at least 60% success in $n$ trials

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#### mathmari

##### Well-known member
MHB Site Helper
Great job rewriting the condition as $a_{10} = \text{Pr}\left(\displaystyle\sum_{i=1}^{10}X_{i}\geq 6\right).$ Your attempt to use Bernoulli's weak law of large numbers is also good. Note, though, that Bernoulli only gives an upper bound for $a_{n}$, so it will not allow us to calculate $a_{10}$ explicitly.

To proceed, expand your previous equation one step further to $$a_{10} = \text{Pr}\left(\sum_{i=1}^{10}X_{i}\geq 6\right) = \text{Pr}\left(\sum_{i=1}^{10}X_{i} = 6\right) + \text{Pr}\left(\sum_{i=1}^{10}X_{i} = 7\right) + \cdots + \text{Pr}\left(\sum_{i=1}^{10}X_{i} = 10\right).$$ To calculate each of the 5 terms on the right, consider using the Binomial Distribution, which calculates the probability of $k$ successes in a sequence of $n$ independent Bernoulli experiments.

Nice work so far. Feel free to let me know if you have any other questions.

Edit: Changed $\text{Pr}(a_{10}\geq0.6)$ to $a_{10}$ because $a_{n}$ is, by definition, the probability of at least 60% success in $n$ trials
I got it!! Now I am able to solve all the questions.. Thank you!!

GJA