Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

1)

A group of six children are choosing colored pencils to draw a picture. Each child is allowed to select one color. The available colors are green, red, and blue. If the second child refuses to use red pencils, and the third child refuses to use blue pencils, then how many ways are there for the children to choose pencils? Assume there are 12 pencils for each color and different children are allowed to choose the same color.

2)

If 20 tickets are sold and two (2) prizes ar to be awarded, find the probability that one (1) person will win both prizes if that person buys exactly two (2) tickets.

3)

X is a normally distributed random variable with a standard deviation of 4.00. Find the mean of x if 12.71% of the area under the distribution curve lies to the right of 14.56. *Please note-I can't get the curve on here, so if you can't help on this one its ok. The answer choices are 13.3, 11.3, 10.0, and 9.5 (wasn't sure if that would help or not). I understand if there is not enough information to answer this.

4)

A school club consist of 20 male students and 15 female students. If 4 students are selected to represent the club in the student government, what is the probability 2 will be female and 2 will be male?

Anything you guys could help with would be great. Thank you

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

Hi Skyblast72

OK. glad you got the other question sorted.

Let's take these one at a time because once you can do one, it may help you do do another without any more help. That means we all know you have progressed with the topic.

1)

A group of six children are choosing colored pencils to draw a picture. Each child is allowed to select one color. The available colors are green, red, and blue. If the second child refuses to use red pencils, and the third child refuses to use blue pencils, then how many ways are there for the children to choose pencils? Assume there are 12 pencils for each color and different children are allowed to choose the same color.

There are enough of each colour so supplies won't run out even if all the children choose the same. That's a big bonus as the question gets much harder if what one child chooses affects what the next may choose!

The first child has 3 choices. {GRB}

The next won't have red, so that child has just 2 choices. {GB}

So far that is 3 x 2 = 6 possibilities {GG, GB, RG, RB, BG, BB}

Carry on like this, multiplying the choices together for all six children.

What do you get?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

I get 324. Thank you I was trying to ad dinstead of multiply. Your forum is very helpful!!

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,456

Hi;

For 2). The only way this problem makes any sense and based on what the other questions are it is a binomial distribution problem.

For 4) I get

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

324 is good.

2)

If 20 tickets are sold and two (2) prizes ar to be awarded, find the probability that one (1) person will win both prizes if that person buys exactly two (2) tickets.

So if there are 20 tickets in a hat and you have just one, what is the probability you'll win?

Let's say that happens.

Now you discover you have a second ticket.

What's the probability that it gets chosen too.

Now, once again multiply the answers, as you want both events to occur.

Bob

ps. bobbym. I see where you have gone with this, but I don't think that was what the questioner intended. Poorly worded, either interpretation.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,456

Hi;

I went for the simplest type of question, I was incorrect in thinking that mine answered that. Your idea looks better to me.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

Sorry for the poor wording, it is exactly how the question is typed on the study guide. I came up with 20C2 = 1 in 190 chance. Is that correct?

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,456

A probability is a ratio or fraction. What did you put in the denominator?

I would have done it like this

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,030

bobbym wrote:

A probability is a ratio or fraction. What did you put in the denominator?

I would have done it like this

Ummm... I don't think that is correct. There are 2 prizes and 20 tickets...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,456

Hi;

First chance is 1 / 20 , second chance is 1 / 19. I am stuck on that.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

You don't need 'combinations at all.

So if there are 20 tickets in a hat and you have just one, what is the probability you'll win?

That'll be 1/20

Let's say that happens.

Now you discover you have a second ticket.

What's the probability that it gets chosen too.

Only 19 tickets left and you have one more ticket to check so that'll be

1/19

Now, once again multiply the answers, as you want both events to occur.

So 1/20 times 1/19

Now for question 3.

Would you like a quick reminder of normal distribution theory? I've got my diagrams all lined up and ready to roll.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,456

Hi anonimnystefy;

You did not read my post. If we treat it as a combinatoric problem as he did then what is the denominator? Where did he go?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

I think the clue that this is just a simple event 1 followed by event 2 probability is the clause "if that person buys exactly two (2) tickets."

As we are told nothing about the other players we cannot assume they all have two tickets (or one or three)

Hence my interpretation.

I think it should say "You have two tickets. What is your chance of winning with both?"

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

Sorry I'm in class. I don't have my notes with me. Will get back to you

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

I am incorrect 1/380 is correct

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

bob bundy wrote:

You don't need 'combinations at all.

So if there are 20 tickets in a hat and you have just one, what is the probability you'll win?

That'll be 1/20

Let's say that happens.

Now you discover you have a second ticket.

What's the probability that it gets chosen too.

Only 19 tickets left and you have one more ticket to check so that'll be

1/19

Now, once again multiply the answers, as you want both events to occur.

So 1/20 times 1/19

Now for question 3.

Would you like a quick reminder of normal distribution theory? I've got my diagrams all lined up and ready to roll.

Bob

Still working on #2 I'm unsure of answer now.

Yes please for question 3 a reminder of normal dustribution

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

hi Skyblast72

Some variables are said to be normally distributed.

eg height of people and IQ may be.

One example which definitely is normal is a bit more complicated to describe but it will help you to understand why it is used.

A sugar packing factory produces 2 lb packs of sugar. In practice, they know that there will be some variation between packs and they don't want complaints from customers who buy a pack, weigh it themselves and find it is underweight. To avoid this they set the packing machine to put 2.1 lbs in each pack, hoping this reduces the chance of an underweight pack.

To check this the quality control people take samples of 10 packs and check the mean weight. These samples will follow a normal distribution. So you can use the theory to check how likely it is that a sample mean will be under 2 lb. Managers have decided to pass the quality check if this probablity is less that 0.001 If the calculation for a particular sample is above 0.001 they call in the engineers to check the machinery.

Diagram 1 shows some typical normal distributions. All have the same symmetrical shape; they differ in two respects (i) the mid point is the mean and this will be different for different situations. (ii) the spread of the curve will vary. This is measured by the standard deviation (= sq rt of variance).

The y axis is set to make the total area under the curve equal to 1.

More to come. Sorry got to go out now.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

**Normal distribution part two.**

To work out a probability using the normal distribution you need the area under the curve.

My second diagram shows the area between x=a and x=b shaded green. This area would give you the probability that x lies between a & b.

It's very hard to get these areas by calculus so computers have been used to generate a table of probabilities.

But every mean and standard deviation would give rise to a differently shaped curve (same basic shape but maybe more squashed up and with the midpoint at a different place).

So how do we get around the impossibility of making an infinite number of tables to cover every problem?

Well there's a simple solution.

If you take the mean away from every x value this shifts the whole curve so that it is centred on x = 0 ( y axis)

I've shown this in my next diagram with a red shift arrow.

Then dividing by the standard deviation converts the amount of spread to a sd of 1. (green squash on the diagram)

The normal distribution tables are for mean = 0 and sd = 1. Every problem can be converted to this, so one table will do all problems.

In practice it is usual to only give half the x values as the other half can be found by symmetry. Also you've got to stop somewhere (have a maximum x)

In theory x could be anythng right up to infinity, but in practice, values beyond about x = 3 are so unlikely they are left out.

The version at

http://www.mathsisfun.com/data/standard … table.html

gives the probabilities from x = 0 up to x = 3. P = 0 up to P = 0.4990

Your own table may give the probabilities in a different way, but you can always adapt a problem by using symmetry and the fact that each half of the graph adds up to P = 0.5

**Question 3.**

X is a normally distributed random variable with a standard deviation of 4.00. Find the mean of x if 12.71% of the area under the distribution curve lies to the right of 14.56. *Please note-I can't get the curve on here, so if you can't help on this one its ok. The answer choices are 13.3, 11.3, 10.0, and 9.5 (wasn't sure if that would help or not). I understand if there is not enough information to answer this.

Let's call the mean 'm'.

The x value in this problem is 14.56 so the standardized conversion is

My final diagram (hint: always draw one of these) shows the area to the right of this x value shaded green. We are told 12.71% of the area lies in this green area so the probability of being there is 0.1271

Now I'm going to use the table mentioned above.

That gives probabilities from x = 0 up to the line rather than beyond it to the right, so I need to do this sum:

It's a great page as you can slide the vertical line across the graph and see the probabilty and standardized x displayed. You can even click the **z onwards** button and see the 12.71 directly!

But without this aid:

I need to hunt for this probability in the table. Got it at 1.14 so finally calculate m from this:

I'll leave you to do this and get m. It is one of the multi choice answers you gave in post 1.

Let me know how you get on with this.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

I haven't forgotten about you just been working on another class, will be back tomorrow. Thanks for all your help.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

Ok. I'll still be here.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

I got m=10

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,836

That's what I got too.

Now for Q4.

4)

A school club consist of 20 male students and 15 female students. If 4 students are selected to represent the club in the student government, what is the probability 2 will be female and 2 will be male?

There are formulas for this but I think you will get a better understanding if you see a diagram.

You've got to choose 4 students and at each choice there are 2 possibiliities for the outcome, M or F.

The diagram is called a Tree Diagram. For choice number one you draw two branches, label one M and the other F

From each end you draw another set of two branches, again marking them M and F, giving four outcomes so far {MM, MF, FM, FF}

Continue like this for two more choices. The final diagram now has 2 x 2 x 2 x 2 = 16 outcomes.

Below I've given you a start with the diagram but I've left some labels and outcomes for you to complete.

You should be able to do

Later edit. This isn't right! See my post 34 for the corrected version.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

I get 16 using the tree diagram

Offline

**Skyblast72****Member**- Registered: 2012-09-30
- Posts: 16

.38 is what I got Thanks

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 108,456

Hi;

I am not getting .38

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline