Mathematics - Ordinary Level - Paper 2, Q.6 - Probability

MGSman123

Probability.

I dropped down to Ordinary level just before my mocks and I still haven't really gotten a full understanding of probability. My main trouble is with 2007, Paper 2, Question 6, (c) parts ii, iii, and iv.

Any help and/or explanation on how to do it? Kinda bewildered...

(It's my backup question anyway)

tootyflutty

MGSman123 wrote: »

Probability.

I dropped down to Ordinary level just before my mocks and I still haven't really gotten a full understanding of probability. My main trouble is with 2007, Paper 2, Question 6, (c) parts ii, iii, and iv.

Any help and/or explanation on how to do it? Kinda bewildered...

(It's my backup question anyway)

Ok when I worked them out, I got for
i) 5x4x3=60
ii) 2x4x3=24
iii) 4x3x1=12
iv) 2x3x1=6

:pac:

Finical

tootyflutty wrote: »

Ok when I worked them out, I got for
i) 5x4x3=60
ii) 2x4x3=24
iii) 4x3x1=12
iv) 2x3x1=6

:pac:

They are all right, I just done that question myself. :cool:

fig mclough

please help, I haven't a clue of paper2!!! the questions I'm gonna try are 1,2,3,7,11!!!

I'm trying to revise but none of the stuff will go in!!
Anyone have any suggestions?? thanks

Dante

Just keep doing exam questions. The only real effective way to study maths IMO. If that fails, id suggest hanging yourself.

MGSman123

tootyflutty wrote: »

Ok when I worked them out, I got for
i) 5x4x3=60
ii) 2x4x3=24
iii) 4x3x1=12
iv) 2x3x1=6

:pac:

Oh, yeah, they are they answers. I'm looking at the marking scheme. What I'm trying to figure out, though, is the process for achieving those answers.

Solas Way

MGSman123 wrote: »

Probability.

I dropped down to Ordinary level just before my mocks and I still haven't really gotten a full understanding of probability. My main trouble is with 2007, Paper 2, Question 6, (c) parts ii, iii, and iv.

Any help and/or explanation on how to do it? Kinda bewildered...

Took me a bit to find the a website to find the exam papers (mine start at 2005, a firend had spares and I dropped down 3 weeks ago><)

Main thing to remember is the numbers in the spaces are the availible options and have no link with the numbers given in the question, (as in 2, 3, 4, 5, 6 in this one). That's what got me for a while.


So I hope this is the question:


How many different three-digit numbers can be formed from the digits
2, 3, 4, 5, 6, if each of the digits can be used only once in each number?

How many of the numbers are less than 400?
5 Numbers, 5 options.
looking at the first box only 2 of those numbers are less than 4. 2 and 3
and it doesn't really matter what we put in the 2nd and 3rd boxes, as the number will still be less than 400.

so 2x4x3=24

How many of the numbers are divisible by 5?
Seeing as it is divisable by 5, the last digit must be either 0 or 5 and since 0 isn't one of the given digits only 5 can go there.
so the third digit has one possibility.
the first two slots don't really matter for this part.
so 4x3x1=12

How many of the numbers are less than 400 and divisible by 5?
Like in part (ii) it must end in 5, The first digit must be less that 4, leaving 2 options and the middle doesn't matter.

2x3x1=6

Hope I did this correctly/ havn't confused you as I'm not the clearest person.
BTW Good luck:)
Hopefully we'll get a nice paper.

We

fig mclough wrote: »

please help, I haven't a clue of paper2!!! the questions I'm gonna try are 1,2,3,7,11!!!

your missing a question

MGSman123

Solas Way wrote: »

Took me a bit to find the a website to find the exam papers (mine start at 2005, a firend had spares and I dropped down 3 weeks ago><)

Main thing to remember is the numbers in the spaces are the availible options and have no link with the numbers given in the question, (as in 2, 3, 4, 5, 6 in this one). That's what got me for a while.


So I hope this is the question:


How many different three-digit numbers can be formed from the digits
2, 3, 4, 5, 6, if each of the digits can be used only once in each number?

How many of the numbers are less than 400?
5 Numbers, 5 options.
looking at the first box only 2 of those numbers are less than 4. 2 and 3
and it doesn't really matter what we put in the 2nd and 3rd boxes, as the number will still be less than 400.

so 2x4x3=24

How many of the numbers are divisible by 5?
Seeing as it is divisable by 5, the last digit must be either 0 or 5 and since 0 isn't one of the given digits only 5 can go there.
so the third digit has one possibility.
the first two slots don't really matter for this part.
so 4x3x1=12

How many of the numbers are less than 400 and divisible by 5?
Like in part (ii) it must end in 5, The first digit must be less that 4, leaving 2 options and the middle doesn't matter.

2x3x1=6

Hope I did this correctly/ havn't confused you as I'm not the clearest person.
BTW Good luck:)
Hopefully we'll get a nice paper.

Wow! Thanks a bunch. That cleared up loads. Only thing I'm wondering is, why did ya pick, for the first part, 4 and 3 at the end?

You know, 2x4x3=24. I kinda get that the 5 in 5! is being replaced by the 2, isn't it, but where are the 1 and 2 gone? (i.e. 5!=5x4x3x2x1)

-MGSman123.

Solas Way

How many different three-digit numbers can be formed from the digits
2, 3, 4, 5, 6, if each of the digits can be used only once in each number?

It's a 3 digit number so we're not asked to go down to 2 and 1. (2x1)

I picked 4 and 3 as the 2nd and 3rd digits because I already used 1 digit for the first one. so only have 4 letters left that could go into slot 2 and then only 3 digits left that could go into slot 3.

A bit late but may come in handy sometime.