Featured 2 weeks ago

See below:

This sets up as a binomial probability. When working with this type of thing, I like to start with this relation:

So we have 50 steps that the drunk is taking, giving

**1**

It turns out that we can find

Therefore we have:

**2**

We're now looking at

To get to

In fact, because he is taking an even number of steps, it is impossible to land on the 50th step on a value of

The probability of landing on

**3**

Here we have

I typed this into google calculator and it came up with an error! Google spreadsheet, however, handled it fine. It gives

Featured 2 weeks ago

See below:

A **permutation** is a method to calculate the number of ways we can take some number of objects and arranging them in some order. For instance, we can say that we have 5 books and we want to put them on a shelf. The number of ways we can arrange those 5 books is

We can also take a smaller grouping and arrange those. For instance, from the 5 books, we want to take a group of 3 of them and arrange them on a shelf. How many ways can we do that? That calculation is:

Note that we can express the two situations above as:

and

So what is **circular permutation**?

Let's say that instead of books, we have 5 people and they are going to sit at a table in a restaurant. What's different here is that instead of being in a line, the people are in a circle, and so having person A sitting in seat 1 with person B in seat 2 on the right and person E on the other side is the same as that same arrangement with person A in seat 2 (with B in seat 3 and E in seat 1). And seat 3, and seat 4, and seat 5.

And so we end up with duplicates of the arrangements. On a circular table, ABCDE is the same as EABCD is the same as DEABC, and so on.

And so to eliminate the duplicates, we divide by the number of places. In this case, we get:

Notice that with

Let's go one step further. We have 5 people who walk into a restaurant but only 3 are going to eat - two people will sit at the bar. In how many different ways can we arrange the 3 people at the table?

We know that in our permutation, we have

We can list those out:

ABC, ACB

ABD, ADB

ABE, AEB

ACD, ADC

ACE, AEC

ADE, AED

BCD, BDC

BCE, BEC

BDE, BED

CDE, CED

Featured 2 weeks ago

1) .20

2)

3)

1) So to begin with you have two scenarios one with mean 10 and the other with mean 20 with some probability of occurrence. The questions are centered around some probability of minutes.

With normal random variables we can think of normal distributions which are symmetrical around the mean. In other words as we go further from the mean the less likely something is to occur. We need to figure out what that mean and spread is aka variance. The variance can be used to determine the probability of observing values further away from the mean.

The normal distribution has some nice theoretical properties and the branch of statistics has found some formulas that tend to hold. One of which is called Expectation or the long run average.

Expected mean is defined as

Expected Variance as

First the mean is

now lest solve for the variance is

The normal distribution can compute the probability of observing values away from the mean using the standard deviation (the square root of the variance). everything within two standard deviations account for 95% of all point that you would observe.

so I would expect that most values have a 95% probability of seeing some value between 4.12 and 19.48.

We can compute a Z score which helps quantify the spread of the standard deviation and compute a probability as a result. There is a precomputed table in the back of most statistic books that allow you to convert a Z score into a probability. I will use this approach to answer the questions. the Z score is defined as

2) The second question is geared toward determining in the traffic is busy or not. Here we actually are interested in determining the likelihood of it being late. There is a nice formula for that too. Bayes decomposed joint probability as

Here we can think of

It makes sense that there is traffic because its past the expected time of 10 minutes. Also I know that its at least .5 since anything greater than 20 accounts for 50% of the probability.

3) for this question we need to compute the difference from 11.8 to 8 and then see what percent of the probability from the zscore is equal to this thus.

This is actually close to 1 standard deviation based on earlier calculation from problem 1. I know statistically that

I know I used some approximation for answering 2 and 3. I personally am okay with it because the nature of probability.

Featured 1 week ago

Let's first notice that this is a combination problem because we don't care in what order we pick the books (if we did, that'd be a permutation problem).

The general formula for a combination is:

To do this, let's first see that we're choosing 2 categories from the population of 3 categories. That's

For each of the categories, there are 3 books we can choose from and we want 1. For each category chosen, then, we have:

Since we have 2 categories chosen, we need to square this:

And so altogether we have:

We can list them out (I'll use A, B, and C; L, M, and N; and W, X, and Y to list out the books):

AL BL CL

AM BM CM

AN BN CN

AW BW CW

AX BX CX

AY BY CY

LW MW NW

LX MX NX

LY MY NY

Featured 1 week ago

Let, **events** that the card selected is an **Ace**

and a **Black-coloured,** resp.

Then, the **Reqd. Prob.** is,

We know that,

For **P(A),** there are

selected in

There are

Similarly,

Note that,

**an ace and black-coloured,** i.e., **a black-coloured** ace.

Out of

Clearly,

From

derived by **Respected Andrea S. !**

**Enjoy Maths.!**

Featured 1 week ago

Note that the order of the CDs does matter, so we are wanting

#""^8P_4 = (8!)/((8-4)!) = 8*7*6*5 = 1680#

In other words, there are

Featured 1 week ago

a. 36,036 ways

b. 278,460 ways

There are 7 boys and 13 girls in the class.

When picking teams, we're looking at combinations (we don't care in what order the players are picked). The general formula is:

**a**

We want a team with 2 boys (from a population of 7) and 6 girls (from a population of 13):

**b**

We want a team of 6 with at least 1 boy and 1 girl. So let's first have 1 guaranteed boy and 1 guaranteed girl:

Now we need to fill in the remaining team. We can pick any of the remaining 18 students to fill in the remaining 4 spots:

Featured 4 days ago

To get a three-of-a-kind, we have 3 of one ordinal, and 2 other cards that are not the same ordinal as the first three or each other.

In all of the calculation, we're using combinations (we don't care about the order of the card draw):

There are 13 ordinals in a deck of cards (Ace, or 1, through 10, and then Jack, Queen, and King). We are picking one of the ordinals:

and from that ordinal we want 3 of the suits (there are 4 suits for each ordinal - Spades, Hearts, Clubs, Diamonds):

We then need to get 2 more cards. They can't be of the same ordinal as the first one or each other (so that's 12 remaining ordinals and we're picking 2). We want one of each:

and now we can calculate this:

Featured 2 days ago

(i)

The number of ways of choosing

#n * (n - 1) * ... * (n - (k-1)) = (n!)/((n-k)!) = ""^nP_k#

Having chosen

Hence if the order did not matter, then the number of ways of choosing

#""^nC_k = (""^nP_k) / (k!) = (n!)/((n-k)! k!)#

So in the given example we have:

**(i)**

The number of ways in which

#""^9C_4 = (9!)/((9-4)! 4!) = (9 * 8 * 7 * 6)/(4 * 3 * 2 * 1) = 3024/24 = 126#

**(ii)**

The number of ways of choosing

#""^4C_2 = (4!)/((4-2)!2!) = (4 * 3) / (2 * 1) = 6#

The number of ways of choosing

#""^3C_1 = (3!)/((3-1)!1!) = (3 * 2) / (2 * 1) = 3#

The number of ways of choosing

#""^2C_1 = (2!)/((2-1)! 1!) = 2 / 1 = 2#

Each of these choices is independent. So the total number of ways of choosing

#""^4C_2 * ""^3C_1 * ""^2C_1 = 6 * 3 * 2 = 36#

**(iii)**

We know that the unrestricted number of ways of choosing

Of those possibilities, any that miss out one or two composers is unacceptable. Let us attempt to count the unacceptable combinations.

If Beethoven is missed, then there are

#""^5C_4 = (5!)/((5-4)!4!) = 5" "# unacceptable combinations.

If Handel is missed, then there are

#""^6C_4 = (6!)/((6-4)! 4!) = 15" "# unacceptable combinations.

If Sibelius is missed, then there are

#""^7C_4 = (7!)/((7-4)! 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 42" "# unacceptable combinations.

Of these unacceptable combinations, there is one that misses two composers - namely Handel and Sibelius, leaving us with the

So the total number of unacceptable combinations is:

#5+15+42-1 = 61#

That leaves the number of acceptable combinations as:

#126-61 = 65#

Featured 3 hours ago

a.

There are 15 balls in total.

The problem we run up against is that depending on what the first marble drawn is, getting a marble of a different colour on the subsequent draws changes. And so this will make for a long-ish answer.

We first look at the first draw and see that the probability of drawing the various colours is:

- White
#=9/15=3/5# - Black
#=4/15# - Green
#=2/15#

**a**

When we pick **White first**, we want the probability of not drawing another white. There are now 14 balls and 8 of them are white, so the probability of first drawing White then Not White is:

When we pick **Black first**, we want the probability of not drawing another black. There are now 14 balls and 3 of them are black, so the probability of first drawing Black then Not Black is:

When we pick **Green first**, we want the probability of not drawing another green. There are now 14 balls and 1 of them is green, so the probability of first drawing Green then Not Green is:

Therefore the probability of drawing 2 balls of different colours is:

**b**

I think the easiest way to work this is to simply show the calculation for each possible series of picks:

**WBG**

**WGB**

**BWG**

and we can stop here. Notice that the denominator isn't changing at all and the terms of the numerator are simply moving around. And so the answer can be found by looking at any one instance and multiplying by 6: