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March 28, 2024, 10:09:08 pm

Author Topic: General Math  (Read 5543 times)  Share 

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roshanajabbour

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General Math
« on: August 03, 2016, 04:28:14 pm »
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Hi, the question is:
A three-digit number is selected from the numbers 3,4,5,7,8,9 with no repetition.
What is the probability that he number formed is greater than 800?

I've already figured out the first part of the answer and found that the total numbers formed was 6P3=120
Can i please get help with finding the rest?
« Last Edit: August 03, 2016, 04:37:17 pm by RuiAce »

RuiAce

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Re: General Math
« Reply #1 on: August 03, 2016, 04:37:35 pm »
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Moderator action was an accident in misclicking. My bad :) overall I edited nothing.

studybuddy7777

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Re: General Math Question Thread
« Reply #2 on: August 03, 2016, 04:44:42 pm »
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Hi I have stickied this topic because we need a new question thread :)
I'll be more than happy to help you answer this question and welcome to atarnotes!

Hi, the question is:
A three-digit number is selected from the numbers 3,4,5,7,8,9 with no repetition
What is the probability that he number formed is greater than 800?


Okay lets get started with this! I'll do it from the start just for clarifications sake :D
We know that anything that starts with a 3,4,5 or 7 will not be bigger than 800. So we can put them aside for now.
834, 835, 837, 839 etc. There are 4 different ways to fill in the third digit.
So for 83? there are 4 possibilities. There are 5 possibilities for 800's (eg 830,840,850)

Thus there are 20 possibilities it can start with an 8. This means that there are also 20 possibilities it can start with a 9. This makes 40 possibilities of it being greater than 800.

But how many possibilities are there? Well there are 20 each for 3, 4, 5, and 7 as well (total 80 possible numbers).

This makes 120 possible numbers. 40 of these are above 800.

Therefore, the possibility is 40/120 or 1/3 (simplifying the fraction).

If I have gone to fast or too slow please tell me so I can adjust my pace. I am currently at the top of my Gen Maths course but am unsure what level you are at. So I just assumed you would want the whole working out.

The short way is 6 x 5 x 4 = 120. (6 ways to fill the 1st number, 5 ways to fill the 2nd number and 4 ways to fill the 3rd number as they are without repetition).

Hope I helped and let me know what method you prefer!

roshanajabbour

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Re: General Math
« Reply #3 on: August 03, 2016, 04:57:48 pm »
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Thank you both! Helped a lot :)

studybuddy7777

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Re: General Math
« Reply #4 on: August 03, 2016, 05:02:20 pm »
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Thank you both! Helped a lot :)


No worries! Feel free to ask any more questions (or any other guests who would like to join/members who would like to post) if you are unsure about how to tackle any question!

My HSC Trial for Gen Maths is tomorrow so i wouldve thought this thread would be a bit more active lol

stephanieazzopardi

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Re: General Math
« Reply #5 on: August 03, 2016, 05:02:57 pm »
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Hi, the question is:
A three-digit number is selected from the numbers 3,4,5,7,8,9 with no repetition.
What is the probability that he number formed is greater than 800?

I've already figured out the first part of the answer and found that the total numbers formed was 6P3=120
Can i please get help with finding the rest?

Just thought I'd post another way of explaining! The best thing to do in this example is to draw a set of boxes and in each box you place the number of possible outcomes at each stage. There will be three boxes in this solution because there are three numbers to be chosen.
|   |  x  |   |  x  |   |

In each box, place the number of items that can be chosen from at each stage.
| 2 |  x  | 5 |  x  | 4 |

Explanation:
Box 1 = 2 because there are 2 possible numbers that can take the place of the first digit in this three digit number. Either 8 or 9 because the number has to be over 800.

Box 2 = 5 because there are 5 possible numbers that can take the place of the second digit in this three digit number. Either 8 or 9 has already been chosen, leaving 5 numbers left that can be chosen.

Box 3 = 4 because there 5 possible numbers that can take the place of the third digit in this three digit number. Because there is to be no repetition in our selection, the 8 or 9 + 1 other digit has already been taken, leaving 4 numbers left than can be chosen

| 2 |  x  | 5 |  x  | 4 | = 40 numbers will be over 800

Therefore, the probability that the number formed will be more than 800 is 40/120 = ⅓. (from provided first answer)

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RuiAce

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Re: General Math
« Reply #6 on: August 03, 2016, 05:53:25 pm »
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Actually I would've written 2x5x4 as well but cause OP used permutation I decided go with the flow

studybuddy7777

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Re: General Math
« Reply #7 on: August 03, 2016, 06:00:44 pm »
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Actually I would've written 2x5x4 as well but cause OP used permutation I decided go with the flow

Just a gentle reminder this is a general maths thread rui :)
But permutations=arrangements yes? Can't say im familiar with the word
Also your formula is really confusing, but looks awesome. Would you mind explaining it? Certainly no offence intended, just need to explain this a bit more to general students ;)

RuiAce

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Re: General Math
« Reply #8 on: August 03, 2016, 06:53:09 pm »
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Just a gentle reminder this is a general maths thread rui :)
But permutations=arrangements yes? Can't say im familiar with the word
Also your formula is really confusing, but looks awesome. Would you mind explaining it? Certainly no offence intended, just need to explain this a bit more to general students ;)
Hm? Well

First number: Two possible numbers to use.

Second number: Five possible numbers to use (because the sixth one is already used)

Third number: Four possible numbers to use (because the fifth one is also already used)

So 2 times 4 times 5


I find it easier to analyse each digit by itself, rather than use permutations.

P.S. Didn't Stephanie already explain it?

studybuddy7777

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Re: General Math
« Reply #9 on: August 03, 2016, 07:15:06 pm »
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Hm? Well

First number: Two possible numbers to use.

Second number: Five possible numbers to use (because the sixth one is already used)

Third number: Four possible numbers to use (because the fifth one is also already used)

So 2 times 4 times 5


I find it easier to analyse each digit by itself, rather than use permutations.

P.S. Didn't Stephanie already explain it?

Ahh ok thanks for that!
P.S. Did Stephanie already explain it? Im not sure but it didnt click with me last time..

Cheers :)

Stefan K

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Re: General Math
« Reply #10 on: October 06, 2016, 11:20:53 pm »
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Guys, I was wondering how to find the mean and sd using a calculator. I completely forgot :(

Stefan K

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Re: General Math
« Reply #11 on: October 07, 2016, 12:05:11 am »
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Hi guys, not sure if they taught you this but my textbook never showed us this (at least I don't think so  ??? ) and there are questions relation to it so here you go :)

jamonwindeyer

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Re: General Math
« Reply #12 on: October 07, 2016, 12:13:19 am »
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Hi guys, not sure if they taught you this but my textbook never showed us this (at least I don't think so  ??? ) and there are questions relation to it so here you go :)

Thanks Stefan! Just remember you won't have to remember anything for the exam that isn't on your formula/reference sheet! ;D

RuiAce

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Re: General Math
« Reply #13 on: October 07, 2016, 12:20:06 am »
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Guys, I was wondering how to find the mean and sd using a calculator. I completely forgot :(
Missed this one. This depends on what calculator you have.

Stefan K

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Re: General Math
« Reply #14 on: October 08, 2016, 10:25:38 pm »
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All good, I got it figured out! :D