##### Questions

##### Question type

Use these controls to find questions to answer

Featured 4 months ago

your friend's contention is incorrect. Instead, a more laborious process of going through potential situations needs to be done.

First thing - we're not talking probability but instead the number of ways something(s) can be arranged (if we were talking probabilities, we'd have a ratio - the number of ways to meet a certain set of conditions divided by the total number of ways the things could be arranged).

The key here is the number of duplicates. There are an unknown number of them - we're told there are "some" but we don't know how many. So let's work with what we do know and see if we can formulate an answer from there.

Your friend has already figured out that we are talking about a combination problem (the general formula of which is:

The reason it's a combination problem is that we don't care about the order in which the books are picked (much like with a poker hand - we don't care how we got the hand but only that we got it).

If we weren't dealing with duplicates at all, we'd have the two people picking 8 books in total from the 40 (non-duplicate) available, which would give us:

So far so good - your friend is on track. So now to deal with the duplicates.

First I'll walk through a simple explanation as to why we're going to do what we're going to do, and then go back to our problem.

When we work with factorials, permutations, and combinations, what we're really doing is starting with a factorial (note the

I'll interweave an example into this discussion. Let's work with a 4-letter word, say ABCD.

A **factorial** takes all the members of a group and arranges them in a unique order. We can make

A **permutation** limits the number of members being arranged. If we only pick 2 letters at a time, we end up with

A **combination** eliminates the differences in the different orders of things - and so AB is the same as BA (much like a poker hand), and so we end up with

To **eliminate duplicates**, we divide by the internal ordering that occurs between said duplicates. So let's say that instead of ABCD we have instead AABC. How many ways can we arrange these letters?

We keep the numerator at

So what happens when we take a population, choose from it, and that population has duplicates. How do we handle it?

Let's take combinations where we pick 2 from AABC. We'll end up with

AA, AB, AC, BC = 4 ways. But if we simply divide by *could* have, we get:

We can expand this thinking to the question at hand, with 40 books in total, 16 potential duplicates, and the picking of 8 books from that population. In short, your friend's contention that

will eliminate *the* duplicate and come up with an accurate tally is incorrect. Instead, a more laborious process of going through potential situations needs to be done.

Featured 3 months ago

Whenever you solve a probability question involving two conditions, and you are being asked to find the probability that either will occur for a given action, you are looking for what is known as a "union probability". Formally speaking, if we say

The trick here is that these two possible events are not disjoint events; in other words, it **can** be possible to pull a single card and have it be a Spade and a Queen at the same time. The formula for determining

(This is read as "the probability of A union B is equal to the probability of A plus the probability of B minus the probability of the intersection of A and B".)

If we consider

If we consider

However, the probability **and** a Queen at the same time. Of all 52 cards in the deck, there is only one Queen of Spades, thus

Thus:

Featured 3 months ago

If the first ball is replaced:

If the first ball is not replaced:

There are

- both are white
- both are black
- one of each color

The probability of one of each color is

If the first ball is replaced:

However, if the first ball is NOT replaced, the probability changes for the second ball. There are fewer of one color and one ball less.

Featured 2 months ago

Only the following two combinations are possible:

I will assume that both the median and the mean are

The median is the value in the middle of a set of data arranged in order. There are

The mode is the value that occurs the most often. The mode is

If the mean is

We already have a total of

Any combination of numbers bigger than 4 will do, except they cannot both be equal to

Only the following two combinations are possible:

The following are not possible:

Featured 2 months ago

This has got to be one of my favourite problems in mathematics...

Let

Let

Now we must use our knowledge of expected outcomes...

Now we can devise a descrete random variable

Getting 0 wins would have a probability of

Getting 1 win is getting a head and then tails,

.

.

.

Hence we must use our

Now factoring out

This stage is the most difficult, we must recognise that:

Now we know

We know:

We can do this as

We can find the derivative using the quotient rule:

We know the total probability is

Featured 4 weeks ago

A few ways of approximating the bell curve...

The general Bell curve:

There are many reasons to why you may want to approximate the bell curve, one being that it is particulaly difficult to integrate definitely...

So by approximating, and finding a function that is more easy to integrate definitely, and has an antiderivative, this is much eaiser.

The first thing I can think of is utilising a function that already has

But by sketching, this doesnt quite have the general shape, so the reciprical could be a good idea:

Now this looks more like it...

One main property of the bell curve, used in statistics regulaly...

This can be proven using polar substitution using the jacobian, a higher level of mathematics....

So one thing we could do, is understand what the total area under

Finding:

We can use the substitution

So

Now use

Now we can use another trig substitution:

Using our substitutions...

Integrating indefinitely...

So

But as from looking from the graph we see that for small values of

So

As it has same value for

Ask a question
Filters

Ã—

Use these controls to find questions to answer

Unanswered

Need double-checking

Practice problems

Conceptual questions