Probability (1 Viewer)

[FDownes]

I'm having a lot of trouble with permutations and combinations, so I was hoping someone here could help me out with a few questions. Please make sure that you explain your working; I really want to understand this topic... The first question asks;

A queue has 4 boys and 4 girls standing in line. Find how mant different arrangements are possible if;

a) The boys and girls alternate.

b) 2 particular girls wish to stand together.

c) All the boys stand together.

d) Also find the probability that 3 particular people will be in the queue together if the queue forms randomly.

EDIT: If it's any help, here are the answers...

a) 1152
b) 10080
c) 2880
d) 3/28

EDIT: Okay, figured out the first two by trial and error. It'd be good if someone could please give me an explanation regardless, so I can understand these problems better.

a) ⁴P₄ x ⁴P₄ x ²P₁ = 1152

b) ⁷P₇ x ²P₂ = 10080

EDIT: I think I understand now. The third one should be;

⁴P₄ x ⁵P₅ = 2880

EDIT: Yup, that's right. But I still don't get the last one.

 

[lolokay]

a) could be boy first or girl first. 4! ways to order the boys, 4! to order the girls, so the answer is 4!*4!*2

b) first treat the 2 boys as one person. There is now 7 people that can be ordered in 7! ways - and 2! ways of ordering the 2 boys

c) treat all the guys as one person. Now you have 5 people - 5! ways of ordering, and 4! of ordering the remaining boys

 

[FDownes]

Thanks, but I'm still stuck on the last one...

 

[lolokay]

d) there are 6 possible groups of 3. There are 8C3 = 56 combinations of 3 boys, so the chance of a particular combination is 6/56 = 3/28

 

[FDownes]

Thanks, now I have a new problem;

A table has 4 boys and 4 girls sitting around it.

a) Find the number of ways of sitting possible if the boys and girls can sit anywhere around the table.

= ⁸P₇
= 5040

b) If the seating is arranged at random, find the probability that;
i) 2 particular girls will sit together.
ii) All the boys will sit together.

 

[sonnylongbottom]

Thanks, now I have a new problem;

A table has 4 boys and 4 girls sitting around it.

a) Find the number of ways of sitting possible if the boys and girls can sit anywhere around the table.

= ⁸P₇
= 5040

b) If the seating is arranged at random, find the probability that;
i) 2 particular girls will sit together.
ii) All the boys will sit together.

b) i. 2! x 6! /5040 = 2/7
ii. 4! x 4! /5040 = 4/35

 

[FDownes]

Thanks very much. Now I have another problem;

A school committee is to be made up of 5 teachers, 4 students and 3 parents.

a) If 12 teachers, 25 students and 7 parents apply to be on the committee, which is chosen at random, how many possible committees could be formed?

This part I've figured out. The answer is 350 658 000.

b) If Jan and her mother both apply, find the probability that both will be chosen for the committee.

This part I'm not sure how to do. The answer is apparently 3036/44275.

 

[lolokay]

b) is just 4/25*3/7 = 12/175. I have no idea where they pulled those numbers from (same answer btw when simplified)

 

[tacogym27101990]

Thanks very much. Now I have another problem;

A school committee is to be made up of 5 teachers, 4 students and 3 parents.

a) If 12 teachers, 25 students and 7 parents apply to be on the committee, which is chosen at random, how many possible committees could be formed?

This part I've figured out. The answer is 350 658 000.

b) If Jan and her mother both apply, find the probability that both will be chosen for the committee.

This part I'm not sure how to do. The answer is apparently 3036/44275.

did you realise 3036/44275 simplifies to 12/175??
cause if thats what you were getting your correct

 

[FDownes]

Well, that explains why I couldn't find the answer. I've got it now, thanks.

 

[FDownes]

Now I have a new problem;

A sample of 3 coins is taken at random from a bag containing 8 ten cent coins and 8 twenty cent coins. Find the probability that a particular ten cent coin will be chosen, if 1 twenty cent and 2 ten cent coins are chosen.

 

[lolokay]

1/4 ?

 

[Forbidden.]

Are these true?

(a) Eighty five percent of lung cancer sufferers have been smokers. Therefore, if we ban
smoking we will reduce the lung cancer deaths by 85%.

(b) A white defendant was identified by a black witness. The defendant’s lawyer claims that
studies show that people make 3 times as many mistakes identifying people of other
races as they make identifying people of their own race. The black witness’s evidence is
therefore not to be trusted.

(c) A certain town has 4 times the expected number of leukemia sufferers. There must therefore be an environmental reason for these extra cases of leukemia.

Source: UNSW MATH1231 Algebra Course notes.

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