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# Another weird maths question watch

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1. In a class of students 1/3 are left handed and 1/8 wear a dental brace.
A student is selected at random.

What is the probability of that student.
a) is left handed AND wears a brace
b) is left handed but does not wear a brace
c) does not wear a brace AND is not left handed

Note: 1/3 and 1/8 DOES NOT = 1 :/

Any ideas? for a) I did : 1/3 * 1/8
2. (Original post by Reversed Flash)
In a class of students 1/3 are left handed and 1/8 wear a dental brace.
A student is selected at random.

What is the probability of that student.
a) is left handed AND wears a brace
b) is left handed but does not wear a brace
c) does not wear a brace AND is not left handed

Note: 1/3 and 1/8 DOES NOT = 1 :/

Any ideas? for a) I did : 1/3 * 1/8
That's correct for a)
for b) and c) think about the fractions which do not wear a brace/are not left handed
3. dld
(Original post by Lemur14)
That's correct for a)
for b) and c) think about the fractions which do not wear a brace/are not left handed
Hey thanks for the answer but the probability dont add up to 1? so how would a be correct?
4. (Original post by Lemur14)
That's correct for a)
for b) and c) think about the fractions which do not wear a brace/are not left handed
Also you got a 9 in maths and 8 in english HOW :/

Please give me tips I am aspiring for those grades this year
5. (Original post by Reversed Flash)
In a class of students 1/3 are left handed and 1/8 wear a dental brace.
A student is selected at random.

What is the probability of that student.
a) is left handed AND wears a brace
b) is left handed but does not wear a brace
c) does not wear a brace AND is not left handed

Note: 1/3 and 1/8 DOES NOT = 1 :/

Any ideas? for a) I did : 1/3 * 1/8
a) 1/24
b/ 7/24
c 14/24 = 7/12
6. (Original post by Party-P3opl3-9)
a) 1/16
b/ 7/16
c 14/16 = 7/8
How did you work it out??
7. (Original post by Reversed Flash)
dld

Hey thanks for the answer but the probability dont add up to 1? so how would a be correct?
The reason why they don't add up because being left handed and having braces are independent from each other. Some people might be right handed and not have braces. The number of people left handed, right handed and both handed would add up to one as they are forgot the word, but you get what I mean. Hope that helped!
8. (Original post by Reversed Flash)
How did you work it out??
You just multiply the probabilities, for example, say that the chance of having a cat was 1/3 and the chance of having a dog is 1/5.

To calculate the chance of having a dog, but not a cat is

2/3 * 1/5 = 2/15

It is 2/3 because 1/3 of people have cats, 1 - 1/3 = 2/3
9. draw a probability tree
___________
I__________ I
LH ________RH
I___ _ I___ I ____I
B ___NB _B ___NB
(Note: B=brace and NB=no brace)

Simply follow the tree down the correct paths and multiply the corresponding probabilities
You can calculate the probabilities for RH and NB because in each line the probability will add to 1, so the probability of being left handed, plus the probability of being right handed = 1.

So for a, you want the path that has 'left handed' and 'brace'. Therefore you are correct in multiplying the probabilities for these 2

for question B, you want to follow the path of LH then NB, multiplying these probabilities.

Edit: Sorry the formatting is a little off so its not a great tree
10. (Original post by Party-P3opl3-9)
You just multiply the probabilities, for example, say that the chance of having a dog was 1/3 and the chance of having a dog is 1/5.

To calculate the chance of having a dog, but not a cat is

2/3 * 1/5 = 2/15

It is 2/3 because 1/3 of people have cats, 1 - 1/3 = 2/3
But how will it work when the probabilities don't add up wot 1?
11. For the first one I think it's 1/3×1/8=1/24

For b is it not just 1/3×7/8=7/24?

And for c I think it is 2/3×7/8=14/24=7/12
12. Hmmmm what's exactly weird here?

The probability of two independent events happening is P(left hand) * P(Dental brace)

Adding up P(left hand)+ P(right hand)+ P(Ambidextrous) will give you 1
Adding up P(Braces) + P(Cast) + P(Band Aids) + P(others) will give you 1
13. (Original post by Lissy14)
draw a probability tree
___________
I__________ I
LH ________RH
I___ _ I___ I ____I
B ___NB _B ___NB
(Note: B=brace and NB=no brace)

Simply follow the tree down the correct paths and multiply the corresponding probabilities
You can calculate the probabilities for RH and NB because in each line the probability will add to 1, so the probability of being left handed, plus the probability of being right handed = 1.

So for a, you want the path that has 'left handed' and 'brace'. Therefore you are correct in multiplying the probabilities for these 2

for question B, you want to follow the path of LH then NB, multiplying these probabilities.

Edit: Sorry the formatting is a little off so its not a great tree
Ah no problem I drew the diagram so for a) it would be 1/24.
14. (Original post by TaintedLight)
Hmmmm what's exactly weird here?

The probability of two independent events happening is P(left hand) * P(Dental brace)

Adding up P(left hand)+ P(right hand)+ P(Ambidextrous) will give you 1
Adding up P(Braces) + P(Cast) + P(Band Aids) + P(others) will give you 1
Just got it thanks! Forgot the No braces and right hand!
15. (Original post by Reversed Flash)
But how will it work when the probabilities don't add up wot 1?
Well, you multiply them (as I have done)

The probabilities only add up to one if it's the same thing, if you get me (not the best at explaining things).

Say if you had your name in a hat, along with 9 other people from the class. The chance of you being picked is 1/10. However, the chance of someone being picked is from your class (i.e everyone but you) is 9/10. 1/10 + 9/10 is 10/10 = 1 (as someone is going to be picked from a hat).

If we said there was a 1/10 chance of you being picked from the hat AND being picked for, I don't know, to take the register back out of the 10 people, the chance of you doing both is 1/10 * 1/10 = 1/100. However, 1/10 + 1/10 =/= 1 as both instances are independent of each other (i.e. there isn't a 100% chance you will be picked for either).
16. (Original post by Party-P3opl3-9)
Well, you multiply them (as I have done)

The probabilities only add up to one if it's the same thing, if you get me (not the best at explaining things).

Say if you had your name in a hat, along with 9 other people from the class. The chance of you being picked is 1/10. However, the chance of someone being picked is from your class (i.e everyone but you) is 9/10. 1/10 + 9/10 is 10/10 = 1 (as someone is going to be picked from a hat).

If we said there was a 1/10 chance of you being picked from the hat AND being picked for, I don't know, to take the register back out of the 10 people, the chance of you doing both is 1/10 * 1/10 = 1/100. However, 1/10 + 1/10 =/= 1 as both instances are independent of each other (i.e. there isn't a 100% chance you will be picked for either).
Just got it thanks! Forgot the No braces and right hand! Very stupid off me. Havnt done maths for ages. thanks again!
17. (Original post by Reversed Flash)
dld

Hey thanks for the answer but the probability dont add up to 1? so how would a be correct?
They don't have to add up to one, as the probabilities are mutually exclusive.
(Original post by Reversed Flash)
Also you got a 9 in maths and 8 in english HOW :/

Please give me tips I am aspiring for those grades this year
I'm not sure tbh...maths just seems to come to me naturally and the English, while I put a lot of work in, was a total shock! Basically you need to listen to your teachers, they will give you the skills and feedback needed to get the high grades, then do plenty of practise questions (make sure you know your equations/quotes!) and apply all your knowledge effectively in the exam
18. ic
(Original post by Lemur14)
They don't have to add up to one, as the probabilities are mutually exclusive.

I'm not sure tbh...maths just seems to come to me naturally and the English, while I put a lot of work in, was a total shock! Basically you need to listen to your teachers, they will give you the skills and feedback needed to get the high grades, then do plenty of practise questions (make sure you know your equations/quotes!) and apply all your knowledge effectively in the exam
For English any specific way you wrote your answers (recipe?) + When did you start revising to get these amazing grades. I am revising moderatly now but alot of people are telling me it is too much?
19. (Original post by Reversed Flash)
ic

For English any specific way you wrote your answers (recipe?) + When did you start revising to get these amazing grades. I am revising moderatly now but alot of people are telling me it is too much?
No, we were actually told we shouldn't keep the same structure throughout our answer as it didn't show sophisciation (and as you can tell I can't spell )
You don't want to hear about my study habits...I got really lucky with my grades given my revision! For a few subjects I did stuff from around Christmas (I'm talking basically the sciences as there's loads of content, and the odd bit of English lit), mostly after school revision sessions ( I did 2 hours of science revision every Friday after school, with at least 1 hour being biology, from the second week in January to May because of problems we'd had with biology teachers meaning we were completely failing biology).
For most exams though, it was a full set of notes made the day before the exam and getting lucky about what comes up which unfortunately can't really be replicated. For my first exam I did loads (wouldn't advise that, although do focus on your earlier exams to begin with), but after that was pretty much whole spec youtube videos from the day before (although I did get a little ahead over half term). Sorry not many useful tips to learn from me!
I'd suggest properly starting around February half term and slowly ramping up to maybe 4 hours a day over exam period maximum

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