Question

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve fax messages, and consider a sample of 25 incoming calls. (Round your answers to three decimal places.)

(a) What is the probability that at most 4 of the calls involve a fax message?


(b) What is the probability that exactly 4 of the calls involve a fax message?

(c) What is the probability that at least 4 of the calls involve a fax message?


(d) What is the probability that more than 4 of the calls involve a fax message?

Answer 1

@mathmale

Answer 2

@zepdrix

Answer 3

Binomial distribution?

Answer 4

Oo boy I actually don't know >.<
Sorry.

Not familiar with probability stuff.
When I first looked at it, I thought I knew, but nope :3 lol

Answer 5

It's okay thanks for trying

Answer 6

I found a solution online. The math is easy I am just trying to understand the concept.

Answer 7

I'm reading about Binomial Distribution on wikipedia...
maybe we can figure this out...
https://en.wikipedia.org/wiki/Binomial_distribution

We want cumulative distribution for the first part, ya?
Because we have to consider 0, 1, 2, 3 or 4 fax messages coming in the 25.

\[\large\rm \Pr(X\le4)=\sum_{i=0}^{4}\left(\begin{matrix}25 \\ \rm i\end{matrix}\right)p^i(1-p)^{25-i}\]Where our probability p=1/4.

Although, if you've already found a solution,
then maybe this just makes it more confusing XD lol

Answer 8

well I am trying to find that formula on my lecture slides. there are several formulas that are very similar

Answer 9

I have this one p(y) = [(n!)/y!(n-y)]*p^y*(q)^(n-y)

Answer 10

Ok good, yes that's what I have in my summation, but with different letters.
Small typo though, the (n-y) should be factorial.

Answer 11

\[\large\rm \Pr(y)=\frac{n!}{y!(n-y)!}p^y q^{n-y}\]Where q=1-p.

Answer 12

I see what q is but what is it? What does it illustrate?

Answer 13

Can you just explain to me what n,y,and q represent?

Answer 14

p represents the given probability of an even happening.
For our problem, that's the 25%.

q is the compliment of that, odds of event NOT happening.
so q=75%

Answer 15

n = total trials

Answer 16

y is a `specific number of successful trials`.

So when we write Pr(y),
we're asking, what are the odds of getting exactly `y` fax messages from the 25 total calls we receive.

Answer 17

So then for part B,
Pr(4) gives us our answer.
That make sense?

That is the probability of getting `exactly` 4 faxes out of 25 calls.

Answer 18

Wait.. so our sample total sample is 100 but we are only drawing 25 from the initial sample, so the compliment is 75 found by q = 100-25 = 75?

Answer 19

maybe I should be more clear:
p = odds of success
q = odds of failure

If odds of success = 25%,
then odds of failure = 100% - 25% = 75%

It's always based off of 100%.
Sometimes you fail, sometimes you succeed,
but `always`, 100% of the time, something happens, right?

So success + failure = 100%

Answer 20

The 25 sample had nothing to do with our p or q.
It was the 25% that was our magic number.

Answer 21

Okay that makes sense.

Answer 22

Let's plug in what we know so far,\[\large\rm \Pr(y)=\frac{25!}{y!(25-y)!}\left(\frac{1}{4}\right)^y \left(\frac{3}{4}\right)^{25-y}\]

Answer 23

I plugged in `25 for every n`,
`1/4 which is equivalent to 25% for our p`,
`3/4 which is equivalent to 75% for our q`.

Answer 24

I get that. So we are going to sum 4 times?

Answer 25

For part A, we want to know,
Pr(0) `odds of getting zero faxes in total`
Pr(1) `odds of getting one fax`
Pr(2) `odds of getting two faxes`
...
Pr(3)
Pr(4)

Yes, we have to add all of those up.

Answer 26

Okay let me work on this for a bit. I will attach an image with my work in a bit..

Answer 27

cool

Answer 28

I am going to be awhile so if you want to help other people then go right ahead.

Answer 29

I'm doing it on paper as well XD lol

Answer 30

You have some options for shortcutting the first fraction if you like.

Answer 31

\[\large\rm \frac{n!}{y!(n-y)!}=\left(\begin{matrix}n \\ \rm y\end{matrix}\right)=nCy\]

Answer 32

We would read this as "n choose y".
You can use your calculator if that's easier for you.

For example, at least with my calculator,
I can type 25 nCr 2 and it gives me 300

That's a shortcut to writing out 25!/2!(25-2)! which will be a little more burdensome.

Answer 33

Whatever works for you though.

Answer 34

is that a quicker way of writing it? The solution I found on line uses the formula you stated up above. I was hesitant to use it because I didn't know where it came from.

Answer 35

Yes, if your teacher hasn't taught you that method yet,
maybe he/she wants you to do it out the long way.
Seems tedious though :P

I'm using a simple TI-30XIIS.
I think I have all of the graphing calculators on my phone/tablet though.
I can probably find the button.

Answer 36

Mmm no, I guess I don't have that fancy beast :)
TI-82, 83, 85 and 86.
Maybe same buttons though, if you need to know where it is.

Answer 37

There has be some quicker way to solving these... I am just going to use nCy

Answer 38

what is C?

Answer 39

How do you use nCy?

Answer 40

Oh oh, ok I think I found the buttons.

You go, `2nd` `Multiply(math in yellow)` `F2 to get PROB` `F3 to get nCr`

Answer 41

But before you press F3 to get your nCr,
fire type `25`,
then press `F3`
then type `0`

That will calculator (25 0) for you, the first one.

Answer 42

hold on here

Answer 43

It will leave the probability window up for you,
so you don't have to navigate through "math" every time

Answer 44

k

Answer 45

25,nCr(0) ?

Answer 46

I found nCr()

Answer 47

Did it give you brackets?
Maybe your calculator wants you to put it in like this then,
nCr(25,0)

Answer 48

mine is a little different maybe :P

Answer 49

when I did nCr(25,1) = 1

Answer 50

You mean nCr(25,0)?

Answer 51

YEAH LOL

Answer 52

nCr(25,1) should give you 25 D:

Answer 53

ok good :)

Answer 54

So we've got a handle on the shortcut, yay

Answer 55

the 1 is illustrating we are starting at 1 then going to 4?

Answer 56

nCr(25,1 <------ ?

Answer 57

Yes,
Your Pr(0) has a nCr(25,0) in it.
Your Pr(1) has a nCr(25,1) in it.
... and so on.

If that's what you were asking.

Answer 58

yes yes that is what I am asking. Awesome

Answer 59

\[\large\rm \Pr(y)=\color{orangered}{\frac{25!}{y!(25-y)!}}\left(\frac{1}{4}\right)^y \left(\frac{3}{4}\right)^{25-y}\]

So this is what we've done to shorten things up for ourselves,\[\large\rm \Pr(y)=\color{orangered}{nCr(25,y)}\left(\frac{1}{4}\right)^y \left(\frac{3}{4}\right)^{25-y}\]

Answer 60

thank you so much and thank you for explaining it to me. I should be fine here on out.

Answer 61

Cool :)
Was good for me to brush up on some of this stuff too haha

Answer 62

I've got one more question

Answer 63

@zepdrix

Answer 64

sup :)

Answer 65

instead of multiplying every nCr(25,1) by each increment can I just input this in my calculator nCr(25,4)*[P^y(1-p)^(n-y)+....]

Answer 66

is that how it works?

Answer 67

is there a function? How does it know r=y?

Answer 68

`n` and `r` are the variables we normally use.
we're using y instead of r, no big deal though.

Answer 69

so everywhere there is a y i will have to use an r...

Answer 70

Hmm I'm not sure what you're asking.
You're asking if you can put it into your calculator as a function?
Or you're asking if you can factor something out of each term?

Answer 71

can i factor nCr out of each term?

Answer 72

No, because they're all different :\
It's nCr(25,0) in the first term,
nCr(25,1) in the next,
and so on.

None of those are the same.

Answer 73

If this is way tedious, you can ... "super shortcut" and use wolfram :) lol

Answer 74

https://www.wolframalpha.com/input/?i=sum+from+y%3D0+to+4+of+%2825+choose+y%29*%281%2F4%29%5E%28y%29*%283%2F4%29%5E%2825-y%29

Answer 75

I can't use my laptop on the exam XD

Answer 76

mmm true true -_-

Answer 77

I thought nCr was a function

Answer 78

I will figure it out, thanks though!