Trig Help? Probability. nCr P^r q^n-r equation?

[lizzy16]

So, I'm doing homework. And, I'm so lost.

Its been months since this unit and I don't understand.

question 26)

a game contains 9 keys. 5 kets have numbers.

4 keys have colors.

Each key is equally likley to get pressed.

this chart which i'm assuming is the key pad to the game.

Red 1 blue

2 3 4

green 5 yellow

Questions:

If 3 keys are pressed at random in the game find the probability of selecting

a) exactley 1 color key

b) at least 2 color keys

c) at most 1 color key

d) All color keys

e) exactley 1 number key

f) at least 1 number key.

Now, i thought I should use the nCr P6r Q^n-r equation.

But, whats n,c, q and q?

And, why can't I just be like..

A) 9 keys. 1 color key. thats 5/9 chance...?

Thats wrong..but why?

[Guest]

So, I'm doing homework. And, I'm so lost.

Its been months since this unit and I don't understand.

question 26)

a game contains 9 keys. 5 kets have numbers.

4 keys have colors.

Each key is equally likley to get pressed.

this chart which i'm assuming is the key pad to the game.

Red 1 blue

2 3 4

green 5 yellow

Questions:

If 3 keys are pressed at random in the game find the probability of selecting

a) exactley 1 color key

b) at least 2 color keys

c) at most 1 color key

d) All color keys

e) exactley 1 number key

f) at least 1 number key.

Now, i thought I should use the nCr P6r Q^n-r equation.

But, whats n,c, q and q?

And, why can't I just be like..

A) 9 keys. 1 color key. thats 5/9 chance...?

Thats wrong..but why?

A) is not a 5/9 chance because first of all, there are 4 color keys and second of all, you have 3 keys pressed. Therefore, the probability of getting only 1 color key out of 3 keys pressed cannot be 5/9. And it's not 4/9 either. 4/9 is the probability that you will press a colored key if you only press ONE key. Make sense?

As far as the equation you gave,

P = the Probability of Success for each individual trial

Q = the Probability of Failure for each individual trial

= 1-P

nCr =The number of different, unordered combinations of r objects from a set of n objects.

= n! / r! (n-r)!

Where

n=the total number of trials.

r=the total number of successful trials.

Okay, let's apply this formula to one of your problems.

First of all - there's a giant difference between EXACTLY and AT LEAST/AT MOST...

EXACTLY is easy. It's just a simple Binomial Probability. So, let's pick one of those EXACTLY questions.

Say, the first one: A.) Exactly 1 color key.

So, in this problem:

P = the probability of getting a color key everytime you press a key = 4/9

Q = the probability of getting a number key instead = 5/9

n = number of keys pressed = 3

r = number of color keys we want out of the 3 keys pressed = 1

So, plugging it into the Binomial Probablity equation:

nCr = 3! / (1! * 2!) = 3

P^r = (4/9)^1 = 4/9 or 0.44

Q^(n-r) = (5/9)^2 = 0.31

And 3*0.44*0.31 = 0.41

So, there is a 41% chance that you are going to get exactly 1 color key if you press 3 keys at random.

Make sense?

Now, here's the trick... AT LEAST or AT MOST is a Cumulative Binomial Probability. Because, you will have to add all the Binomial Probabilities of each possible successful outcome.

So, let's take the 2nd problem... AT LEAST 2 color keys.

The successful outcomes are:

1.) exactly 2 color keys

2.) exactly 3 color keys

So, you need to solve the Binomial Probability of each one of those successful outcomes (just like I showed you on the first problem) and then add the results together.

Okay, I'm just showing you here the process. But, you really have to understand WHY we do it this way. It will take me too long to explain to you why and I'm sure your book explains it much better than I can anyway...

Hope this helps some.

Edited May 5, 2011 by anatess

[MarginOfError]

I'm hoping I can make a more illustrative demonstration of what is taking place here. If I fail, sorry.

So we're calling our experiment the selection of three keys from a set of nine. We are also stipulating that the selection happens simultaneously. That is to say, we aren't selecting them one at a time, but rather, selecting the group as a whole. (that may not be important practically, but mathematically it is a very important distinction).

One of the things that characterizes our experiment is that the objects we are selecting have two distinct characteristics. So each object either is a color or a number. For the purposes of our experiment, we are going to think of them as either a color or not a color.

Since we have four colored keys, we can say that the probability of a single randomly selected key being colored is 4/9. Let's call this p

Since we have five non-colored keys, we can say that the probability of a single randomly selected key being non-colored is 5/9 = (9-4)/9. Let's call this q

Now, for (A), we want to know the probability of selecting exactly 1 colored key and 2 non-colored keys. So the set would look like this

(colored) (non-colored) (non-colored)

The probability of any such set matching set is

(4/9) * (5/9) * (5/9)

Now remember your mathematical properties, and write this with exponents

(4/9)^1 * (5/9)^2

Coincidentally, this is starting to look a bit like the binomial probability equation. Well, actually, not so coincidentally, but you can see it forming.

p^1 * q^2

But here's the catch. The last expression only give the probability of one combination of 1 colored and 2 non-colored. So we need to find out how many such combination exist. In this example, we can easily count them--there's one for each of the four colors:

(4/9)^1 * (5/9)^2 +

(4/9)^1 * (5/9)^2 +

(4/9)^1 * (5/9)^2 +

(4/9)^1 * (5/9)^2 + = 4 * (4/9)^1 * (5/9)^2

We can simplify this by using the nCr function where n = 4, r = 1

4C1 * (4/9) ^ 1 * (5/9) ^ 2 = 0.412

###############################################

For part B, if we want the probability of at least two keys being colored, we are actually asking

Pr(two colored keys) + Pr(three colored keys)

So we apply the formula again

(4C2 * (4/9)^2 * (5/9)^1) + (4C3 * (4/9)^3 * (5/9)^0)

= 0.329 + 0.088

= 0.417

I hope that's enough to get you visualizing the thought process.

[lizzy16]

Thank you both :) I figured it out. I've already learned this and was just confused trying to remember.

But, the refreshers helped me find the right anwseres.

[rex8499]

I really really really hated statistics. I'd much rather re-do differential equations than have to take even a simple statistics class.

[pam]

This thread just reminds me of all the reasons I avoided math as much as possible. :)

[slamjet]

[Guest]

I would rather take Math classes than English classes. Heck, I would rather take Math classes than Pilipino classes. Come to think about it, I'd rather take Math classes than any class including Homeroom!

[pam]

I would rather take Math classes than English classes. Heck, I would rather take Math classes than Pilipino classes. Come to think about it, I'd rather take Math classes than any class including Homeroom!

Give me any class to do with history and I'm in seventh heaven.

[Guest]

Give me any class to do with history and I'm in seventh heaven.

LOL! History is one of my most hated subjects.

Test Question:

1.) Who was the leader of the Katipunan?

Answer: How should I know! Whoever he is, he died long time ago.

[pam]

LOL! History is one of my most hated subjects.

Test Question:

1.) Who was the leader of the Katipunan?

Answer: How should I know! Whoever he is, he died long time ago.

haha I don't know who the leader is. I just know the Katipunan is a revolutionary group in the Philippines.

[Guest]

haha I don't know who the leader is. I just know the Katipunan is a revolutionary group in the Philippines.

Oh yeah, and another Test Question:

1.) When did Magellan land in the Philippines?

Answer: A long time ago. It was so long ago all the great grand children of those who were alive then are all dead now.

It just irritated me how I had to memorize all those stuff. Memorization is something I'm just very weak on (which is why I'm struggling in my calling of Primary Song Leader - I'm asking the kids to memorize the songs when I, myself can't remember the words!). And I always think that it doesn't matter one wit if I remember what day it happened and what are the names of the people involved - as long as I remember that it happened and why it happened!

[Dravin]

I would rather take Math classes than English classes. Heck, I would rather take Math classes than Pilipino classes. Come to think about it, I'd rather take Math classes than any class including Homeroom!

I'd rather take math, inasmuch as at this point no math class has required I write a paper.

[Wingnut]

LOL! History is one of my most hated subjects.

Test Question:

1.) Who was the leader of the Katipunan?

Answer: How should I know! Whoever he is, he died long time ago.

Actual Facebook status yesterday from a 14-year-old young women in the ward I share a building with:

I honestly don't think I'm gonna do very well in History this year since I've had to ask my teacher both who is Ronald Reagan and Osama Bin Laden is

Did I mention that she attends a prestigious and private all-girls school?

[Guest]

I'd rather take math, inasmuch as at this point no math class has required I write a paper.

All the math classes in my master's program except for Statistics required a paper. I'm glad to be finished with it!

[Guest]

Actual Facebook status yesterday from a 14-year-old young women in the ward I share a building with:

Did I mention that she attends a prestigious and private all-girls school?

Yeah, reminds me of that Kelly-somebody country singer.

[Dravin]

All the math classes in my master's program except for Statistics required a paper. I'm glad to be finished with it!

I'm undergraduate so things are easier for me. I have to take (if the requirements at IUPUI are equivalent to those at UVU) Calc I, Calc II and Principles of Statistics then I'm done with my Math requirements.