# Make the internet a better place to learn

3

Parzival S.
Featured 2 weeks ago

See below:

#### Explanation:

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

${\sum}_{k = 0}^{n} {C}_{n , k} {\left(p\right)}^{k} {\left(1 - p\right)}^{n - k} = 1$

So we have 50 steps that the drunk is taking, giving $n = 50$. We also have that a step forward has probability of $p = .4$. All we need now is $k$.

1

It turns out that we can find $k$ by looking at where the man ends up. If over the course of 50 steps he ends up at $x = - 10$, we know he's taken 30 steps backwards and 20 steps forwards - which gives $k = 20$. We don't care about the series of steps he took (whether FBFBBBFBFF... or whatever) because the combination term of the binomial will find all the possible ways of making those steps.

Therefore we have:

${C}_{50 , 20} {\left(.4\right)}^{20} {\left(.6\right)}^{30} \approx 0.1146$

2

We're now looking at $x = - 27$

To get to $- 27$, our drunk needs to have taken some number of steps to get there. For instance, let's say our drunk for the first 48 steps goes back and forth between 1 and 0. Now he has 2 steps left to take. Can he end up on $x = 1$? No.

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

The probability of landing on $x = - 27$ on the 50th step is 0.

3

Here we have $n = 2000 , p = 0.01$, and we're looking at $k = 23$:

${C}_{2000 , 23} {\left(.01\right)}^{23} {\left(.99\right)}^{1977}$

I typed this into google calculator and it came up with an error! Google spreadsheet, however, handled it fine. It gives $\approx 0.0671$

1

## What is circular permutation? give some examples also

Parzival S.
Featured 2 weeks ago

See below:

#### Explanation:

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 5! = 120 ways.

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:

$5 \times 4 \times 3 = 60$ different ways. The permutation formula allows us to find that number directly:

P_(n,k)=(n!)/((n-k)!); n="population", k="picks"

Note that we can express the two situations above as:

P_(5,5)=(5!)/(0!)=5! = 120

and

P_(5,3)=(5!)/(2!)=(5xx4xx3xx2!)/(2!)=5xx4xx3=60

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:

(5!)/5=120/5=24

Notice that with 5! = 5xx4xx3xx2xx1, we can also express this as:

(5!)/5=4! = 24

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 (5!)/(2!)=3! = 60, so 60 ways to arrange the 5 people in groups of 3 at the table. We then need to work with the table arrangement and recognize that ABC is the same as CAB, so we divide by 3 - the number of seats at the table, to get 20.

We can list those out:

ABC, ACB
ABE, AEB
ACE, AEC
BCD, BDC
BCE, BEC
BDE, BED
CDE, CED

2

## How can i calculate the probability of waiting for a certain bus to come? Difficult but interesting

Will
Featured 2 weeks ago

1) .20
2)$\approx .81$
3) $\approx .34$

#### Explanation:

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 $\setminus \sum p \left(x\right) \cdot x$

Expected Variance as $\setminus \sum p \left(x\right) \cdot {\left(x - \setminus \mu\right)}^{2}$

First the mean is $.18 \cdot 20 + .82 \cdot 10 = 11.8$
now lest solve for the variance is $.18 \cdot {\left(20 - 11.8\right)}^{2} + .82 \cdot {\left(10 - 11.8\right)}^{2} = 14.76$

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.
$\setminus \sqrt{14.76} \approx 3.84$

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 $\frac{x - \setminus \mu}{\setminus} \sigma$ and the final probability equates to $.2$

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 $p \left(a | b\right) = \frac{p \left(b | a\right) p \left(a\right)}{p \left(b\right)}$

Here we can think of $b = x > 15$ and $a = l a t e$ so we have $p \left(b\right) = .2$, $p \left(a\right) = .18$ for $p \left(b | a\right)$ i use the same standard deviation from above and mean as 20 for my distribution. Here I make a loose assumption that this is reasonable estimate
$\frac{.9 \cdot .18}{.20} = .81$

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.

$11.8 - 8 = 3.8$

This is actually close to 1 standard deviation based on earlier calculation from problem 1. I know statistically that $\pm 3.84$ is about 68% so i can take about half of that to approximate roughly this amount. Thus 34% of it coming between now and 15 minutes.

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

1

## I have three distinct mystery novels, three distinct fantasy novels, and three distinct biographies. I'm going on vacation, and I want to take two books of different genres. How many possible pairs can I choose?

Parzival S.
Featured 1 week ago

$\left(\begin{matrix}3 \\ 2\end{matrix}\right) {\left(\begin{matrix}3 \\ 1\end{matrix}\right)}^{2} = 3 \times {3}^{2} = 27$

#### Explanation:

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:

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

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

$\left(\begin{matrix}3 \\ 2\end{matrix}\right)$

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

$\left(\begin{matrix}3 \\ 1\end{matrix}\right)$

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

${\left(\begin{matrix}3 \\ 1\end{matrix}\right)}^{2}$

And so altogether we have:

$\left(\begin{matrix}3 \\ 2\end{matrix}\right) {\left(\begin{matrix}3 \\ 1\end{matrix}\right)}^{2} = 3 \times {3}^{2} = 27$

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

1

## From a normal pack of 52 playing cards, one card is selected at random. How can I find the probability of the card being either a black card or an ace?

Ratnaker Mehta
Featured 1 week ago

$\frac{7}{13}$.

#### Explanation:

Let, $A \mathmr{and} B$ denote, the events that the card selected is an Ace

and a Black-coloured, resp.

Then, the Reqd. Prob. is, $P \left(A \cup B\right)$.

We know that, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right) \ldots \ldots \left(\star\right)$.

For P(A), there are $52$ cards in a normal pack, out which $1$ can be

selected in $52$ ways.

There are $4$ aces in a pack, so, $1$ can be chosen in $4$ ways.

$\therefore P \left(A\right) = \frac{4}{52.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}_{1}\right)$.

Similarly, $P \left(B\right) = \frac{26}{52.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}_{2}\right)$.

Note that, $A \cap B$ denotes the event that the selected card is

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

Out of $52$ cards, there are $2$ black-coloured aces.

Clearly, $P \left(A \cap B\right) = \frac{2}{52.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}_{3}\right)$.

From $\left(\star\right) , \left({\star}_{1}\right) , \left({\star}_{2}\right) \mathmr{and} \left({\star}_{3}\right)$, we have,

$\text{The Reqd. Prob.} = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$, as already

derived by Respected Andrea S. !

Enjoy Maths.!

3

## Find the number of ways to listen to four CDs from a selection of 8 CDs.? A. 1680 B. 336 C. 70 D. 50

George C.
Featured 1 week ago

""^8P_4 = 1680, i.e. choice A.

#### Explanation:

Note that the order of the CDs does matter, so we are wanting ""^8P_4 rather than ""^8C_4 ...

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

In other words, there are $8$ choices for the first CD, $7$ for the second, $6$ for the third and $5$ for the fourth.

1

## There are 20 students in a class with 7 boys . (a) In how many different ways can a team of 2 boys and 6 girls formed ? (b) In how many different ways can a team of 6 be formed such that there is a least one boy and at least one girl in the team ?

Parzival S.
Featured 1 week ago

a. 36,036 ways
b. 278,460 ways

#### Explanation:

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:

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

a

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

$\left(\begin{matrix}7 \\ 2\end{matrix}\right) \left(\begin{matrix}13 \\ 6\end{matrix}\right) = 21 \times 1716 = 36036$ ways

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:

$\left(\begin{matrix}7 \\ 1\end{matrix}\right) \left(\begin{matrix}13 \\ 1\end{matrix}\right)$

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:

$\left(\begin{matrix}7 \\ 1\end{matrix}\right) \left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}18 \\ 4\end{matrix}\right) = 7 \times 13 \times 3060 = 278460$ ways

1

## There's a deck of 52 cards. A five-card hand has three of a kind consists of 3 cards of the same rank, one card of another, and one card of another. How many different 3-of-a-kind hands are there?

Parzival S.
Featured 4 days ago

$\left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 3\end{matrix}\right) \left(\begin{matrix}12 \\ 2\end{matrix}\right) {\left(\begin{matrix}4 \\ 1\end{matrix}\right)}^{2} = 54 , 912$

#### Explanation:

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):

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

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:

$\left(\begin{matrix}13 \\ 1\end{matrix}\right)$

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

$\left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 3\end{matrix}\right)$

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:

$\left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 3\end{matrix}\right) \left(\begin{matrix}12 \\ 2\end{matrix}\right) {\left(\begin{matrix}4 \\ 1\end{matrix}\right)}^{2}$

and now we can calculate this:

$13 \times 4 \times 66 \times {4}^{2} = 54 , 912$

1

## Can someone help me with this PROBABILITY question? Please.

George C.
Featured 2 days ago

(i) $126$; (ii) $36$; (iii) $65$

#### Explanation:

The number of ways of choosing $k$ items from $n$ distinct items if the order of the choice matters is:

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

Having chosen $k$ items from $n$, then the number of ways in which those $k$ items can be ordered is k!.

Hence if the order did not matter, then the number of ways of choosing $k$ items from $n$ distinct items is:

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

So in the given example we have:

(i)

The number of ways in which $4$ pieces can be chosen from $9$ is:

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

(ii)

The number of ways of choosing $2$ pieces by Beethoven is:

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

The number of ways of choosing $1$ piece by Handel is:

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

The number of ways of choosing $1$ piece by Sibelius is:

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

Each of these choices is independent. So the total number of ways of choosing $4$ pieces as required is their product:

${\text{^4C_2 * ""^3C_1 * }}^{2} {C}_{1} = 6 \cdot 3 \cdot 2 = 36$

(iii)

We know that the unrestricted number of ways of choosing $4$ pieces from $9$ is ""^9C_4 = 126

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 $5$ other pieces, giving us:

$\text{^5C_4 = (5!)/((5-4)!4!) = 5" }$ unacceptable combinations.

If Handel is missed, then there are $6$ other pieces, giving us:

$\text{^6C_4 = (6!)/((6-4)! 4!) = 15" }$ unacceptable combinations.

If Sibelius is missed, then there are $7$ other pieces, giving us:

$\text{^7C_4 = (7!)/((7-4)! 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 42" }$ unacceptable combinations.

Of these unacceptable combinations, there is one that misses two composers - namely Handel and Sibelius, leaving us with the $4$ pieces by Beethoven. That possibility is included twice.

So the total number of unacceptable combinations is:

$5 + 15 + 42 - 1 = 61$

That leaves the number of acceptable combinations as:

$126 - 61 = 65$

2

## A bag contains 15 balls. 9 white , 4 black and 2 green. a)We pick 2 balls. Whats the probability of the balls to be different colors? b)We pick 3 balls. Whats the probability of the balls to be different colors?

Parzival S.
Featured 3 hours ago

a. $\approx 0.6810$; b. $\approx 0.1582$

#### Explanation:

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 $= \frac{9}{15} = \frac{3}{5}$
• Black $= \frac{4}{15}$
• Green $= \frac{2}{15}$

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:

$\frac{3}{5} \times \frac{6}{14} = \frac{18}{60} = \frac{63}{210}$

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:

$\frac{4}{15} \times \frac{11}{14} = \frac{44}{210}$

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:

$\frac{2}{15} \times \frac{13}{14} = \frac{26}{210}$

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

$\frac{63}{210} + \frac{44}{210} + \frac{26}{210} = \frac{143}{210} \approx 0.6810$

b

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

WBG

$\frac{9}{15} \times \frac{4}{14} \times \frac{2}{13} = \frac{72}{2730}$

WGB

$\frac{9}{15} \times \frac{2}{14} \times \frac{4}{13} = \frac{72}{2730}$

BWG

$\frac{4}{15} \times \frac{9}{14} \times \frac{2}{13} = \frac{72}{2730}$

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:

$6 \times \frac{72}{2730} = \frac{72}{455} \approx 0.1582$