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Featured 4 months ago

321 using a single copy of each number.

A whole number is a number that sits within the range

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

1-digit = 1

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

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

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

And now we do the 5-digit numbers:

And now we add them up:

Featured 3 months ago

We use our good friend Bayes' rule to help us deduce:

**Warning: Long answer ahead!**

One form of Bayes' rule states the following:

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:

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

We actually have that here; the

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

From hereon, let's use the shorthand

Now, we do the calculations!

Finally, we place these values back into the equation for

For completeness, *had* to take either 1, 2, or 3 rolls.

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

Featured 3 months ago

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:

and now let's evaluate this:

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:

and now let's evaluate that:

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

And now we can find the probability:

Featured 2 months ago

595

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:

**Adding up:**

0 Women

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

1 Women

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

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

which gives us:

2 Women

There are 8 women that can go into 2 slots:

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

which gives us:

So all told:

**Subtracting down**

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

4 Women

There are 8 women that can go into 4 slots:

3 Women

There are 8 women that can go into 3 slots:

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

which gives us

So all told:

Featured 1 month ago

Mean:

Variance:

Standard deviation:

We are given that

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 = E[X] = sum_("allÂ " x)[x * P(X=x)]#

In our case, this works out to be:

#mu = [0*P(0)]+[1*P(1)]+[2*P(2)]+[3*P(3)]#

#color(white)mu=(0)(0.15)+(1)(0.35)+(2)(0.45)+(3)(0.05)#

#color(white)mu="Â Â Â Â Â Â Â "0"Â Â Â Â Â Â Â "+"Â Â Â Â "0.35"Â Â Â Â "+"Â Â Â Â Â "0.9"Â Â Â Â Â "+"Â Â Â Â "0.15#

#color(white)mu=1.4#

So, over a large number of rolls, we would expect the average roll value to be

The variance is a measure of the "spread" of

That's because the variance *squared* distance between each possible value and

#sigma^2="Var"(X)=E[(X-mu)^2]#

Using a bit of algebra and probability theory, this becomes

#sigma^2=E[X^2]-mu^2#

#color(white)(sigma^2)=sum_("allÂ x")x^2P(X=x)" "-" "mu^2#

For this problem, we get

#sigma^2=[0^2*P(0)]+[1^2*P(1)]+[2^2*P(2)]#

#color(white)(sigma^2=)+[3^2*P(3)]" "-" "1.4^2#

#color(white)(sigma^2)=(0)(0.15)+(1)(0.35)+(4)(0.45)+(9)(0.05)#

#color(white)(sigma^2=)-1.96#

#color(white)(sigma^2)=0.64#

So the average squared distance between each possible

Standard deviation is easyâ€”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 *square* of the units of

That's where standard deviation comes in. The standard deviation

#sigma= sqrt (sigma^2)#

For this problem, that works out to be

#sigma = sqrt(0.64)=0.8#

So every time we pick an *(See: confidence intervals.)*

Featured 1 month ago

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

We want our hand to have exactly one Ace. Since there are four aces in a deck, we can set

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

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

~~~~~

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

Let's evaluate it!

And so:

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