Math methods unit probability help please

[Zahta]

porbability help


In how many ways can a set of 4 grls be selected from 10
a. if the edldest girl is included in each set,
b. if the eldest girl is excluded from each set



Another question

A class consits of 15 students of whom 5  are prefects . How many committees of 8 can be formed , each containing
a. exactly 2 prefects
b. at least 2 prefects

[Anon123]

I still don't get why unit 1/2 prob is harder then unit 3/4 prob

god I miss not being able to do prob

[Zahta]

is unit 1 & 2 prob harder then 3& 4

[Anon123]

is unit 1 & 2 prob harder then 3& 4

Yes. Much harder. Unit 3/4 prob is INSANELY EASY, I kid you not, this is the one thing I can pretty much get full marks on every time - and I am probably just above average, unit 1/2 prob was much, much harder

[geoff_821]

is unit 1 & 2 prob harder then 3& 4

Yes. Much harder. Unit 3/4 prob is INSANELY EASY, I kid you not, this is the one thing I can pretty much get full marks on every time - and I am probably just above average, unit 1/2 prob was much, much harder

Maybe it was hard coz u were learning it for the first time.
Unit 3/4 probability is just a slight extension of 1/2

[Anon123]

Maybe it was hard coz u were learning it for the first time.
Unit 3/4 probability is just a slight extension of 1/2

I dunno, I think I would still fail 1/2 prob if i did it right now...

I mean stuff like E(x), var(x), the normal distribution, markov and all that are just so easy...combinatrics last year was insane

[b^3]

is unit 1 & 2 prob harder then 3& 4

Yes. Much harder. Unit 3/4 prob is INSANELY EASY, I kid you not, this is the one thing I can pretty much get full marks on every time - and I am probably just above average, unit 1/2 prob was much, much harder

Maybe it was hard coz u were learning it for the first time.
Unit 3/4 probability is just a slight extension of 1/2

Not entirely, there are a fair few things in unit 1/2 prob that you don't do in unit 3/4 e.g. combinatrics and all that rubbish. Basically you need basic probability from previous years, binomial + a bit of markov chains plus. Then there is the new stuff that has mean/median e.t.c - the itnergration stuff.

[REBORN]

I love transition matrices or markov chains? is there a difference?

[Aegis]

is unit 1 & 2 prob harder then 3& 4

Yes. Much harder. Unit 3/4 prob is INSANELY EASY, I kid you not, this is the one thing I can pretty much get full marks on every time - and I am probably just above average, unit 1/2 prob was much, much harder

Maybe it was hard coz u were learning it for the first time.
Unit 3/4 probability is just a slight extension of 1/2

Not entirely, there are a fair few things in unit 1/2 prob that you don't do in unit 3/4 e.g. combinatrics and all that rubbish. Basically you need basic probability from previous years, binomial + a bit of markov chains plus. Then there is the new stuff that has mean/median e.t.c - the itnergration stuff.

must know how to draw a tree diagram and use a tree diagram in unit 3/4 prob.

I love transition matrices or markov chains? is there a difference?

I think they are the same but with different names?

[Zahta]

lool thanks guys for the heads up anyone wanna help out with the questions

[MeLucky]

I love transition matrices or markov chains? is there a difference?

Markov chains is the process of finding a pr of something after n steps. But yeah, when I talk about markov chains I usually just say transition matrices.

[kamil9876]

porbability help


In how many ways can a set of 4 grls be selected from 10
a. if the edldest girl is included in each set,
b. if the eldest girl is excluded from each set



Another question

A class consits of 15 students of whom 5  are prefects . How many committees of 8 can be formed , each containing
a. exactly 2 prefects
b. at least 2 prefects

after skimming through I think no one actually answered this, or did we decide it is irrelevant for year12 and hence not going to bother? either way:

a) There are different ways to select the 4 girls in such a way (since the eldest is already there you just need 3 of the remaining 9).

b) So essentially there are 9 girls and you want 4 of them. So .

a) So we have 10 non-prefects and 5 prefects. There are different ways of choosing 6 non-prefects and ways of choosing 2 prefects. Hence in total there are committtees(basically each committee is just a pair of two prefects and six non-prefects)

b) Easier to calculate the compliment, there are committees in total. Find how many contain exactly 1 prefect and how many contain no prefects (use similair reasoning to part a). Now just subtract that from the total.