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## A whole number consists of the 2 4, 6, 8, and 0. How many whole numbers are possible?

Parzival S.
Featured 4 months ago

321 using a single copy of each number.

#### Explanation:

A whole number is a number that sits within the range $0 - \infty$.

We can make whole numbers using one, some, or all of the numbers listed. I'm going to assume that we can only use a single copy of each number, so the number $24$ is ok but numbers $22$ and $44$ are not. So let's see what we can make.

Before we start, I'm noticing that if we make a 2-digit number starting with $0$ - we end up having a single digit number. The same thing is true for a 3-digit number starting with $0$ - we end up with a 2-digit number. So I'm going to skip single digit numbers and instead use 2-digit numbers starting with $0$ to be those single digit numbers. In fact, the only number that we need to count at the single digit level is the number $0$, so let's count that now:

1-digit = 1

For 2-digit numbers, we have a permutation of five numbers, select 2:

#P_(5,2)=(5!)/((5-2)!)=(5!)/(3!)=(5xx4xx3!)/(3!)=20#

For 3-digit numbers, we do the same for five numbers, pick three:

#P_(5,3)=(5!)/((5-3)!)=(5!)/(2!)=(5xx4xx3xx2!)/(2!)=60#

For 4-digit numbers, we do the same for five numbers, pick four:

#P_(5,4)=(5!)/((5-4)!)=(5!)/(1!)=(5xx4xx3xx2xx1!)/(1!)=120#

And now we do the 5-digit numbers:

#P_(5,5)=(5!)/((5-5)!)=(5!)/(0!)=(5xx4xx3xx2xx1!)/(1)=120#

And now we add them up:

$1 + 20 + 60 + 120 + 120 = 321$

## Find the Probability?

Geoff K.
Featured 3 months ago

We use our good friend Bayes' rule to help us deduce:
$P r \left(\text{took 2 rolls"|"score is 5}\right) = \frac{2}{15.}$

#### Explanation:

One form of Bayes' rule states the following:

$P r \left(B | A\right) = \frac{P r \left(A B\right)}{P r \left(A\right)} = \frac{P r \left(A | B\right) \cdot P r \left(B\right)}{P r \left(A\right)}$

This allows us to write a conditional probability of "B given A" in terms of "A given B", which may be easier to calculate. In this question,

B = "die rolled twice", and
A = "score is 5".

Let's write this into the equation:

#Pr("2 rolls"|"score is 5")=(Pr("score is 5"|"2 rolls")*Pr("2 rolls"))/(Pr("score is 5"))#

The numerator on the RHS is easy to deduce; however, the denominator, in its current state, is not. But we can make it easier. We recall that, if $A$ can be partitioned into $k$ disjoint events $A \cap {C}_{1} , A \cap {C}_{2} , \ldots , A \cap {C}_{k}$ (where none of these intersected regions overlap, and together they all form $A$), then

$P r \left(A\right) = {\sum}_{i = 1}^{k} P r \left(A | {C}_{i}\right) \cdot P r \left({C}_{i}\right)$

We actually have that here; the ${C}_{i}$'s will be the number of possible rolls on a turn. What we're saying is that

$P r \left(\text{score is 5")=Pr("score is 5"|"1 roll")*Pr("1 roll}\right)$
#color(white)"XXXXXXXXX"+Pr("score is 5"|"2 rolls")*Pr("2 rolls")#
#color(white)"XXXXXXXXX"+Pr("score is 5"|"3 rolls")*Pr("3 rolls")#

Kind of a "the whole is equal to the sum of its parts" thing.

From hereon, let's use the shorthand ${S}_{5}$ for "the score was 5" and ${R}_{n}$ for "$n$ number of rolls". Putting this all together, we have

$P r \left({R}_{2} | {S}_{5}\right) = \frac{P r \left({S}_{5} | {R}_{2}\right) P r \left({R}_{2}\right)}{P r \left({S}_{5} | {R}_{1}\right) P r \left({R}_{1}\right) + P r \left({S}_{5} | {R}_{2}\right) P r \left({R}_{2}\right) + P r \left({S}_{5} | {R}_{3}\right) P r \left({R}_{3}\right)}$
$= \frac{P r \left({S}_{5} \cap {R}_{2}\right)}{P r \left({S}_{5} \cap {R}_{1}\right) + P r \left({S}_{5} \cap {R}_{2}\right) + P r \left({S}_{5} \cap {R}_{3}\right)}$

Now, we do the calculations!

$P r \left({S}_{5} \cap {R}_{1}\right) = P r \left(\text{roll a 5}\right)$
$= \frac{1}{6}$
$P r \left({S}_{5} \cap {R}_{2}\right) = P r \left(\text{roll a 2, then a 3}\right) = \frac{1}{6} \cdot \frac{1}{6}$
$= \frac{1}{36}$
$P r \left({S}_{5} \cap {R}_{3}\right) = P r \left[\text{roll a 1; then (1,3), (2,2), or (3,1)}\right] = \frac{1}{6} \cdot \frac{3}{36}$
$= \frac{1}{72}$

Finally, we place these values back into the equation for $P r \left({R}_{2} | {S}_{5}\right) :$

$P r \left({R}_{2} | {S}_{5}\right) = \frac{\frac{1}{36}}{\frac{1}{6} + \frac{1}{36} + \frac{1}{72}} \textcolor{b l u e}{\cdot \frac{72}{72}}$

$\textcolor{w h i t e}{P r \left({R}_{2} | {S}_{5}\right)} = \frac{2}{12 + 2 + 1}$

$\textcolor{w h i t e}{P r \left({R}_{2} | {S}_{5}\right)} = \frac{2}{15}$

For completeness, $P r \left({R}_{1} | {S}_{5}\right) = \frac{12}{15} = \frac{4}{5} ,$ and $P r \left({R}_{3} | {S}_{5}\right) = \frac{1}{15.}$ These three probabilities sum to 1, which is what we'd expect, since if the player's score was 5, it had to take either 1, 2, or 3 rolls.

## Bonus:

There's a great Numberphile video on YouTube discussing Bayes' rule here:

## A 12 member jury for a criminal case will be selected from a pool of 15 men and 15 women. What is the probability that the jury will have 6 men and 6 women?

Parzival S.
Featured 3 months ago

$\cong .2896$

#### Explanation:

To find this, we need to know the number of ways the jury can be picked overall, and then the number of ways 6 men and 6 women can be on the jury. The fraction of the two will be the probability.

The total number of ways the jury can be picked is the combination of a pool of 30 people and choosing 12:

#C_(30,12)=(30!)/((12!)(30-12)!)=(30!)/((12!)(18!))#

and now let's evaluate this:

#(30xx29xx28xx27xx26xx25xx24xx23xx22xx21xx20xx19xxcancel(18!))/(12xx11xx10xx9xx8xx7xx6xx5xx4xx3xx2xxcancel(18!))#

$\frac{\textcolor{red}{\cancel{30}} \times 29 \times \textcolor{b l u e}{\cancel{28}} \times \textcolor{g r e e n}{\cancel{27}} \times {\textcolor{\tan}{\cancel{26}}}^{13} \times 25 \times \textcolor{b r o w n}{\cancel{24}} \times 23 \times {\textcolor{\mathmr{and} a n \ge}{\cancel{22}}}^{\textcolor{\tan}{\cancel{2}}} \times 21 \times {\textcolor{p u r p \le}{\cancel{20}}}^{\textcolor{\tan}{\cancel{2}}} \times 19}{\textcolor{b r o w n}{\cancel{12}} \times \textcolor{\mathmr{and} a n \ge}{\cancel{11}} \times \textcolor{p u r p \le}{\cancel{10}} \times \textcolor{g r e e n}{\cancel{9}} \times \textcolor{\tan}{\cancel{8}} \times \textcolor{b l u e}{\cancel{7}} \times \textcolor{red}{\cancel{6 \times 5}} \times \textcolor{b l u e}{\cancel{4}} \times \textcolor{g r e e n}{\cancel{3}} \times \textcolor{b r o w n}{\cancel{2}}}$

$29 \times 13 \times 25 \times 23 \times 21 \times 19 = 86 , 493 , 225$

Ok - we know the number of ways the jury can be picked. Now how many ways can the jury consist of 6 men and 6 women?

For the men, there are 15 in the pool and we're picking 6:

#C_(15,6)=(15!)/((6!)(15-6)!)=(15!)/((6!)(9!))#

and now let's evaluate that:

#(15xx14xx13xx12xx11xx10xxcancel(9!))/(6xx5xx4xx3xx2xxcancel(9!))#

$\frac{\textcolor{red}{\cancel{15}} \times {\textcolor{g r e e n}{\cancel{14}}}^{7} \times 13 \times \textcolor{b l u e}{\cancel{12}} \times 11 \times {\textcolor{g r e e n}{\cancel{10}}}^{5}}{\textcolor{b l u e}{\cancel{6}} \times \textcolor{red}{\cancel{5}} \times \textcolor{g r e e n}{\cancel{4}} \times \textcolor{red}{\cancel{3}} \times \textcolor{b l u e}{\cancel{2}}}$

$7 \times 13 \times 11 \times 5 = 5005$

The same number of choices of women are available, so we multiply the number of men's choices and women's choices:

$5005 \times 5005 = 25 , 050 , 025$

And now we can find the probability:

$\frac{25 , 050 , 025}{86 , 493 , 225} \cong .2896$

## In how many ways can a committee of 4 be selected from 6 men and 8 women if the committee must contain at most 2 women?

Parzival S.
Featured 2 months ago

595

#### Explanation:

We have a committee that will have 4 people and at most 2 can be women. There are 6 men and 8 women that can be on the committee. How many ways can we select the 4 people?

There are two ways we can do this - we can figure out the number of ways the committee can be formed with 0, 1, and 2 women and add them up. The other way to do it is to figure out how many different ways the committee can be formed in all cases, then subtract out the ways that involve 3 and 4 women. Let's do it both ways to show that the answer will be the same.

All of these calculations will be Combinations (we don't care about the order of the picks, just the members of the committee). The general formula for a combination is to look at P, the population (or available number of people who can sit on the committee) and k, the number selected), with the general formula being:

#C_(P,k)=(P!)/((k!)(P-k)!)#

0 Women

There are 6 men and 4 slots, so we have a combination of:

#C_(6,4)=(6!)/((4!)(2!))=(6xx5xx4!)/(2xx4!)=15#

1 Women

There are 8 women that can go into the 1 slot:

#C_(8,1)=(8!)/((1!)(7!))=(8xx7!)/(7!)=8#

and 6 men who can go into the remaining 3 slots:

#C_(6,3)=(6!)/((3!)(3!))=(6xx5xx4xx3!)/(3xx2xx3!)=20#

which gives us:

$8 \times 20 = 160$

2 Women

There are 8 women that can go into 2 slots:

#C_(8,2)=(8!)/((2!)(6!))=(8xx7xx6!)/(2xx6!)=28#

and 6 men who can go into the remaining 2 slots:

#C_(6,2)=(6!)/((2!)(4!))=15# (see the 0 women category for the calculation)

which gives us:

$28 \times 15 = 420$

So all told:

$15 + 160 + 420 = 595$

Subtracting down

The total number of ways we can fill the slots on the committee is:

#C_(14,4)=(14!)/((4!)(10!))=(cancel(14)^7xx13xxcancel(12)xx11xx10!)/(cancel(4xx3)xxcancel(2)xx10!)=7xx13xx11=1001#

4 Women

There are 8 women that can go into 4 slots:

#C_(8,4)=(8!)/((4!)(4!))=(cancel(8)^2xx7xxcancel6xx5xx4!)/(cancel4xxcancel(3xx2)xx4!)=70#

3 Women

There are 8 women that can go into 3 slots:

#C_(8,3)=(8!)/((3!)(5!))=(8xx7xx6xx5!)/(3xx2xx5!)=56#

and 6 men who can go into the remaining 1 slot:

#C_(6,1)=(6!)/((1!)(5!))=(6xx5!)/(5!)=6#

which gives us

$56 \times 6 = 336$

So all told:

$1001 - 70 - 336 = 595$

## How do you find the mean of the random variable #x#?

Geoff K.
Featured 1 month ago

Mean: $\mu = 1.4$
Variance: ${\sigma}^{2} = 0.64$
Standard deviation: $\sigma = 0.8$

#### Explanation:

We are given that $X$ could take on the values $\left\{0 , 1 , 2 , 3\right\}$ with respective probabilities $\left\{0.15 , 0.35 , 0.45 , 0.05\right\}$. Since $X$ is discrete, we can imagine $X$ as a 4-sided die that's been weighted so that it lands on "0" 15% of the time, "1" 35% of the time, etc.

The question is, when we roll this die once, what value should we expect to get? Or perhaps, if we roll the die a huge number of times, what should the average value of all those rolls be?

Well, of the 100% of the rolls, 15% should be "0", 35% should be "1", 45% should be "2", and 5% should be "3". If we add all these together, we'll have what's known as a weighted average.

In fact, if we placed these relative weights at their matching points on a number line, the point that would "balance the scale" is the mean that we seek.

This is a good way to interpret the mean of a discrete random variable. Mathematically, the mean $\mu$ is the sum of all the possible values, weighted by their probabilities. As a formula, this is:

$\mu = E \left[X\right] = {\sum}_{\text{all } x} \left[x \cdot P \left(X = x\right)\right]$

In our case, this works out to be:

$\mu = \left[0 \cdot P \left(0\right)\right] + \left[1 \cdot P \left(1\right)\right] + \left[2 \cdot P \left(2\right)\right] + \left[3 \cdot P \left(3\right)\right]$
$\textcolor{w h i t e}{\mu} = \left(0\right) \left(0.15\right) + \left(1\right) \left(0.35\right) + \left(2\right) \left(0.45\right) + \left(3\right) \left(0.05\right)$
$\textcolor{w h i t e}{\mu} = \text{ "0" "+" "0.35" "+" "0.9" "+" } 0.15$
$\textcolor{w h i t e}{\mu} = 1.4$

So, over a large number of rolls, we would expect the average roll value to be $\mu = 1.4$.

The variance is a measure of the "spread" of $X$. Going back to our "balanced number line" idea, if we moved our weights out from our "centre of gravity" $\mu$ so that they are twice as far away, $\mu$ itself wouldn't change, but the variance would increase, by a factor of 4.

That's because the variance ${\sigma}^{2}$ of a random variable is the average squared distance between each possible value and $\mu$. (We square the distances so that they're all positive.) As a formula, this is:

${\sigma}^{2} = \text{Var} \left(X\right) = E \left[{\left(X - \mu\right)}^{2}\right]$

Using a bit of algebra and probability theory, this becomes

${\sigma}^{2} = E \left[{X}^{2}\right] - {\mu}^{2}$
#color(white)(sigma^2)=sum_("all x")x^2P(X=x)" "-" "mu^2#

For this problem, we get

${\sigma}^{2} = \left[{0}^{2} \cdot P \left(0\right)\right] + \left[{1}^{2} \cdot P \left(1\right)\right] + \left[{2}^{2} \cdot P \left(2\right)\right]$
$\textcolor{w h i t e}{{\sigma}^{2} =} + \left[{3}^{2} \cdot P \left(3\right)\right] \text{ "-" } {1.4}^{2}$
$\textcolor{w h i t e}{{\sigma}^{2}} = \left(0\right) \left(0.15\right) + \left(1\right) \left(0.35\right) + \left(4\right) \left(0.45\right) + \left(9\right) \left(0.05\right)$
$\textcolor{w h i t e}{{\sigma}^{2} =} - 1.96$
$\textcolor{w h i t e}{{\sigma}^{2}} = 0.64$

So the average squared distance between each possible $X$ value and $\mu$ is ${\sigma}^{2} = 0.64$.

Standard deviation is easy—it's just the square root of the variance. But, why bother with it if it's pretty much the same? Because the units of ${\sigma}^{2}$ are the square of the units of $X$. If $X$ measures time, for example, its variance is in units of ${\text{(time)}}^{2}$, which really doesn't help us if we're trying to establish a "margin of error".

That's where standard deviation comes in. The standard deviation $\sigma$ of $X$ is a measure of how far from $\mu$ we should expect $X$ to be. It's simply

$\sigma = \sqrt{{\sigma}^{2}}$

For this problem, that works out to be

$\sigma = \sqrt{0.64} = 0.8$

So every time we pick an $X$, the expected distance between $\mu$ and that $X$ is $\sigma = 0.8$. And since $\sigma$ is in the same "units" as $X$, it's much more easy to use to help us construct a margin of error. (See: confidence intervals.)

## What is the probability that the deal of a five-card hand provides exactly one ace?

Parzival S.
Featured 1 month ago

#P=(778,320)/(2,598,960)~=29.9%#

#### Explanation:

An alternative way to do this is to use equations for combinations, the general formula for which is:

#C_(n,k)=(n!)/((k)!(n-k)!)# with $n = \text{population", k="picks}$

We want our hand to have exactly one Ace. Since there are four aces in a deck, we can set $n = 4 , k = 1$, and so:

${C}_{4 , 1}$

But we also need to account for the other four cards in our hand. There are 48 cards to pick from, so we can set $n = 48 , k = 4$, and so:

${C}_{48 , 4}$

We multiply them together to find the total number of ways we can get exactly one Ace in our hand:

#C_(4,1)xxC_(48,4)=(4!)/((1)!(4-1)!)(48!)/((4)!(48-4)!)=>#

#(cancel(4!)(48!))/((3!)cancel(4!)(44!))=(cancel(48)^8xx47xx46xx45xxcancel44!)/(cancel(3xx2)xxcancel44!)=>#

$8 \times 47 \times 46 \times 45 = 778 , 320$

~~~~~

To figure out the probability, we also need to know the total number of 5-card hands possible:

#C_(52,5)=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))#

Let's evaluate it!

#(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960#

And so:

#P=(778,320)/(2,598,960)~=29.9%#

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