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Rolling 20 times a fair dice

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Hey!! :eek:

Tim rolls $20$ times a fair dice.

  • Calculate the probability to get a "$3$" exactly $8$ times.
  • The probability to get at least $k$ times a "$5$" is equal to $10,18\%$ Calculate $k$.
  • Explain the relation to the term $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$.



I have done the following:

  • The probability to get a "$3$" is equal to $\frac{1}{6}$. If we throw $20$ times the dice the probability to get exactly eight "$3$" is equal to $\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (1-\frac{1}{6}\right )^{20-8}=\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (\frac{5}{6}\right )^{12}\approx 0.84\%$.

    Is this correct?(Wondering)

    $$ $$
  • We have that \begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X\leq k)=10,18\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} How can we continue to get the value of $k$ ? (Wondering)
    $$ $$
  • Is this the formula we used in the previous question, just with other numbers? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,686
Hey!! :eek:

Tim rolls $20$ times a fair dice.

  • Calculate the probability to get a "$3$" exactly $8$ times.
  • The probability to get at least $k$ times a "$5$" is equal to $10,18\%$ Calculate $k$.
  • Explain the relation to the term $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$.

I have done the following:

  • The probability to get a "$3$" is equal to $\frac{1}{6}$. If we throw $20$ times the dice the probability to get exactly eight "$3$" is equal to $\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (1-\frac{1}{6}\right )^{20-8}=\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (\frac{5}{6}\right )^{12}\approx 0.84\%$.

    Is this correct?
Hey mathmari !!

Yep. (Nod)

  • We have that \begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X\leq k)=10,18\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} How can we continue to get the value of $k$ ?
Shouldn't it be $P(X\geq k)=10.18\% \Rightarrow 1-P(X{\color{red}<} k)=10.18\%$ ? (Worried)

Can't we calculate those terms until we get to the percentage? (Wondering)

Alternatively we can use some tool or table to calculate the inverse cumulative binomial function. (Nerd)

  • Is this the formula we used in the previous question, just with other numbers? (Wondering)
Close, but not exactly.
You effectively wrote that that "at least $k$ times a $5$" is the same as $1$ minus "less than $k$ times a $5$".
Or:
$$P(\text{at least k times 5}) = 1- P(\text{less than k times a 5})$$

The value $B\left (10;\frac{5}{6};i\right )$ is the probability that $i$ times out of $10$ we do not roll a $5$.
In our case "at least $k$ times a $5$" is the same as "less than $10-k$ times not a $5$".
In formula form:
$$P(\text{at least k times 5}) = P(\text{less than 10-k times not a 5})$$
(Thinking)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Shouldn't it be $P(X\geq k)=10.18\% \Rightarrow 1-P(X{\color{red}<} k)=10.18\%$ ? (Worried)

Can't we calculate those terms until we get to the percentage? (Wondering)

Alternatively we can use some tool or table to calculate the inverse cumulative binomial function. (Nerd)
So we have the following:
\begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X< k)=10,18\% \\ & \Rightarrow P(X<k)=89,82\% \\ & \Rightarrow P(X\leq k-1)=89,82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} or not? (Wondering)

Could you explain to me further how we could calculate $k$ from here? I got stuck right now. (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Close, but not exactly.
You effectively wrote that that "at least $k$ times a $5$" is the same as $1$ minus "less than $k$ times a $5$".
Or:
$$P(\text{at least k times 5}) = 1- P(\text{less than k times a 5})$$

The value $B\left (10;\frac{5}{6};i\right )$ is the probability that $i$ times out of $10$ we do not roll a $5$.
In our case "at least $k$ times a $5$" is the same as "less than $10-k$ times not a $5$".
In formula form:
$$P(\text{at least k times 5}) = P(\text{less than 10-k times not a 5})$$
(Thinking)
So, do we have that $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ time out of $10$ we do not roll a $5$, i.e. at least twice out of $10$ we don't roll a $5$, i.e. at most $8$ times out of $10$ we roll a $5$ ? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,686
So we have the following:
\begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X< k)=10,18\% \\ & \Rightarrow P(X<k)=89,82\% \\ & \Rightarrow P(X\leq k-1)=89,82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} or not? (Wondering)

Could you explain to me further how we could calculate $k$ from here? I got stuck right now. (Wondering)
Yes.
We can write:
$$\sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}
=\overbrace{\left (\frac{5}{6}\right )^{20}
+ \binom{20}{1}\cdot \left (\frac{1}{6}\right )^1\cdot \left (\frac{5}{6}\right )^{19}
+\ldots}^{k\text{ terms}}
\approx\overbrace{0.26\% + \ldots}^{k\text{ terms}}
=89.82\%$$
We can approximate each term... it is quite a few though... (Sweating)

Perhaps we should take the other approach. (Thinking)

So, do we have that $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ time out of $10$ we do not roll a $5$, i.e. at least twice out of $10$ we don't roll a $5$, i.e. at most $8$ times out of $10$ we roll a $5$ ?
You are off by $1$ in each case. (Worried)
The sum is up to and including $i=2$, isn't it?
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Yes.
We can write:
$$\sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}
=\overbrace{\left (\frac{5}{6}\right )^{20}
+ \binom{20}{1}\cdot \left (\frac{1}{6}\right )^1\cdot \left (\frac{5}{6}\right )^{19}
+\ldots}^{k\text{ terms}}
\approx\overbrace{0.26\% + \ldots}^{k\text{ terms}}
=89.82\%$$
We can approximate each term... it is quite a few though... (Sweating)

Perhaps we should take the other approach. (Thinking)
Do you mean such a table ? (Wondering)


You are off by $1$ in each case. (Worried)
The sum is up to and including $i=2$, isn't it?
Ah yes, so is it as follows?

$\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ nor $2$ time out of $10$ we do not roll a $5$, i.e. at least three times out of $10$ we don't roll a $5$, i.e. at most $7$ times out of $10$ we roll a $5$

(Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,686
Do you mean such a table ? (Wondering)
Such a table yes, except that that one is for the t-distribution instead of the binomial distribution.

How about this table?

Ah yes, so is it as follows?

$\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ nor $2$ time out of $10$ we do not roll a $5$
Shouldn't it be:

$\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ nor $2$ times out of $10$ we do not roll a $5$, ... (Wondering)