Statistics Help please!: Permuations and Combinations:)

Hey guys. Would you mind helping me with the following please: It's for revision, not homework so I would appreciate as much help as possible please :smile:

Also I have done some, would you mind seeing if I have done them right please? Thanks...

1. 7 men and 5 women have to be nominated for a committee which contains 4 members.

How many ways can the committee be chosen? I did 12C4=495

How many ways will the committee consist of 2 men and 2 women? I did 7C2x5C2=210.

Find the probability the commitee will contain exactly two men. I did 210/495x100=42.42%

2. 10 discs. 5 have A on them, 3 have B on them and 2 have C on them. They are placed in a bag and seleted at random and placed in a straight line. The sequence of 10 makes a code word.

How many different possible code words can be made? I did 10!/2!x3!x5!= 2520.

How many different possible code words start and end with the letter B? I did 8! all over 10!/2!= 1/45

Find the probability that a code word starts and finishes with the same letter. How do I do this?

3. 6 cards have letters A-F on them. They are shuffled and placed in a row.

How many different arrangements are there? I did 6!=720

How many of these arrrangements are the vowels next to each other? I did 5x2!=10.

The cards are placed face down. 3 of the cards are selected at random. Find the probability at least one of the cards is a vowel. How do I do this?

4. 5 bricks, 1 blue, 1 green, 1 yellow, 2 white.

How many different arrangements are there? I did 5!/2!=60

Find the probability the two white are at the end of the line. I did 3! all over 5!/2!= 1/10

5. Letters CCEGGGGEE give a tune.

How many different tunes can be formed? I did 9!/2!x3!x4!=1260.

How many of these tunes have the Gs in position 2,4,6,8. I did 4x4!/2!=48

Find the probability that the Gs will be in this position. I did 48/1260x100=3.8%.

Thanks for your help and time :smile:

Reply 1

TheCurlyHairedDude

How is that going to help anyone in the real world lol ¬¬

Reply 2

ghostwalker

Original post by Robpattinsonxxx

1. 7 men and 5 women have to be nominated for a committee which contains 4 members.

How many ways can the committee be chosen? I did 12C4=495

How many ways will the committee consist of 2 men and 2 women? I did 7C2x5C2=210.

Find the probability the commitee will contain exactly two men. I did 210/495x100=42.42%

Yep.

2. 10 discs. 5 have A on them, 3 have B on them and 2 have C on them. They are placed in a bag and seleted at random and placed in a straight line. The sequence of 10 makes a code word.

How many different possible code words can be made? I did 10!/2!x3!x5!= 2520.

Yep.

How many different possible code words start and end with the letter B? I did 8! all over 10!/2!= 1/45

Not sure what you've done there.
If you place a B at either end, then you're left with 5As, 1B, 2Cs, and the number of arrangements of those will be found using the same process as the first part.

Find the probability that a code word starts and finishes with the same letter. How do I do this?

Well when you've done the previous part, you'll have the number that start and end with a B, so you now need to work out the number that start and end with an A, and the number that start and end with a C, and adding them will give the number that start and end with the same letter.

3. 6 cards have letters A-F on them. They are shuffled and placed in a row.

How many different arrangements are there? I did 6!=720

Yep.

How many of these arrrangements are the vowels next to each other? I did 5x2!=10.

Since the vowels are next to each other, you can treat them as if they are one item (with the caveat that item can arrise in two ways AE or EA). So how many ways can you arrange 5 items and then multiply by 2 for the two possible ways the vowels can be arranged.

The cards are placed face down. 3 of the cards are selected at random. Find the probability at least one of the cards is a vowel. How do I do this?

Consider the complementary event, that none of the cards are vowels and then subtract your probability from "1".

4. 5 bricks, 1 blue, 1 green, 1 yellow, 2 white.

How many different arrangements are there? I did 5!/2!=60

Find the probability the two white are at the end of the line. I did 3! all over 5!/2!= 1/10

Nope - this uses the same methodology as question 2.

5. Letters CCEGGGGEE give a tune.

How many different tunes can be formed? I did 9!/2!x3!x4!=1260.

Yep.

How many of these tunes have the Gs in position 2,4,6,8. I did 4x4!/2!=48

Nope. Once the Gs are placed, then you are left with 2Cs and 3Es to place in 5 slots. Same methodology as before, as if you were putting the 5 items in a row (the fact that the slots are not side by side is irrelevant).

Find the probability that the Gs will be in this position. I did 48/1260x100=3.8%.

Thanks for your help and time :smile:

Method is correct, but the "48" is wrong from previous part.

Reply 3

Robpattinsonxxx

Original post by ghostwalker

Yep.

Yep.

Not sure what you've done there.
If you place a B at either end, then you're left with 5As, 1B, 2Cs, and the number of arrangements of those will be found using the same process as the first part. So 8!/5!x2!

Well when you've done the previous part, you'll have the number that start and end with a B, so you now need to work out the number that start and end with an A, and the number that start and end with a C, and adding them will give the number that start and end with the same letter.
I get 1344

Yep.

Since the vowels are next to each other, you can treat them as if they are one item (with the caveat that item can arrise in two ways AE or EA). So how many ways can you arrange 5 items and then multiply by 2 for the two possible ways the vowels can be arranged.So its not 5!x2! or 5x2! but 5!x2?

Consider the complementary event, that none of the cards are vowels and then subtract your probability from "1". 1-(2/3)^3=0.7037. This right?

Nope - this uses the same methodology as question 2.3!=6

Yep.

Nope. Once the Gs are placed, then you are left with 2Cs and 3Es to place in 5 slots. Same methodology as before, as if you were putting the 5 items in a row (the fact that the slots are not side by side is irrelevant). 5!/3!x2!=10

Method is correct, but the "48" is wrong from previous part.

10/1260x100

Reply 4

ghostwalker

Original post by Robpattinsonxxx

Note: I can't quote a quote.

So 8!/5!x2!
Yes.

1344.
No.

Its best to think of it as 5!x2! (since then generalizing to 3 items together gives the same format)
Note: 5!x2! = 5!x2

Consider the complementary event, that none of the cards are vowels and then subtract your probability from "1". 1-(2/3)^3=0.7037. This right?

No.
What's the probatility that the first one is not a vowel?
Then once you've chosen that, what about the second? The situation has now changed.

5!/3!x2!=10
Yep, and subsequent.

Reply 5

Robpattinsonxxx

Is it 4/6 x 3/5 = 2/5 for the vowel one? Why have we done it as not picking the vowels, would it be 3/5?

Reply 6

ghostwalker

Original post by Robpattinsonxxx

Is it 4/6 x 3/5 = 2/5 for the vowel one? Why have we done it as not picking the vowels, would it be 3/5?

You're on the right lines. You've only done the calculations for picking two cards, whereas the question asks for three (at least according to your original posting).

It's done this way, because otherwise you'd have to work out the probability of 1 vowel, and the probability of 2 vowels, and add them. You could do it that may, and it would probably be good for you to do so, just for the practise and you can check your answer against the other method; you will notice however, that it is more involved.