Need Help for The Probability Question (Still not convincing!!!!) (2 Viewers)

[bikapika]

No offence but i really find the probability question strange.
i think it should be:

Ways for 0 red 6 yellow: 12C6=924..................................................(1)
Ways for 1 red 5 yellow : 12C1 X 12C5 X 6P6/(5!) = 57024..............(2)
since you can withdraw the 6 marbles in any order (6P6 occurs) but some yellows are repeated ( 5! Occurs)
Ways for 2 red 4 yellow: 12C2 X 12C4 X 6P6/(2!4!) = 490050………….(3)
Ways for 3 red 3 yellow: 12C3 X 12C3 X 6P6/(3!3!) = 968000………….(4)

Now [(1) + (2) + (4)] X 2 + (3) = 2063996…….(5)
in order to take account of 0 yellow 6 red, 1 yellow 5 red ……..
Note (3) is left out from multiplication as 3Y3R and 3R3Y are repeated (Thanks to Mujoo)

So for part (i), you go:

(4) / (5) = 0.47

For part (ii), you go, since you have the same ways to select 0 yellow 6 red as (1), 1 yellow 5 red as (2)... .etc

[(3) + (2) + (1)]/ (5) = 0.27

So since this approach produces a different answer from A.Morris and Terry Lee's method (which I still don’t find it convincing) What do you think?



Please tell me where i did wrong although i will only get 70ish but i really want to work this question out since probability is my favour

 

[bikapika]

A.Morris and Terry Lee reckons:

i.) 12C3 X 12C3 / 24C6 =0.36
ii.) (12C4x12C2+12C5x12C1 + 12C6)/24C6=0.32

What im uncertain about this method is the fact that the order of withdrawal is not taken into account.

A simple 12C3 X 12C3 as "possible ways of selection" does not consider whether you can withdraw in different orders, for example,

RRRYYY, which has 12 x 11 x 10 x 12 x 11 x 10 /(3!3!)ways to select............(1)
R"Y"RRYYY, which has 12 x 12 x 11 x 10 x 11 x 10/(3!3!) ways to select........(2)
.....etc

However (1) and (2) are mutually exclusive ways to withdraw the marbles. So a simple 12C3 x 12C3 as E(events) only takes account of (1) OR (2) and so on, but not the sum of (1) and (2).

Similarly, when you calculate the S(Sample space) you will have the same problem with a simple 24C6

As result, I find this method inconvinsive and hence decide to develop another method which i find more reasonable -0- omg ppl please reply

 

[Mooju]

Initially, that's what I thought of, but when you look at the question, order isn't important. Think about it this way, when you grab 3 marbles right? of any colour, is the order you pick them out important if you're concerned with what the final result is, it's exactly like the question in the exam. You've got 12 marbles, 6 red and 6 yellow, and the question is concerning itself with 3 reds and 3 yellows, not to exactly have 2 reds, a yellow a red and then the other 2 yellows if you catch my meaning. It is concerned with a GROUP of 3 reds and 3 yellows not the ORDER of 3 reds and 3 yellows. That's why I still agree with Terry's method in doing it.

Plus also what my teacher said at one point. Always ask yourself if order is important. Generally that'll tell you whether you're dealing with perms or combs. In this case about puicking out and worried about just have how many balls of what in the end result, I concluded it was combinations.

If THAT didn't make sense I'm sorry

XD

Really lousy when it comes to explaining probability and stuff --> it gets confusing T_T

 

[joshuajspence]

I hate probability. I don't think anyone got the same answer as me so I probably did it wrong.

For part A, I got

(12 * 11 * 10 * 12 * 11 * 10) / (24 * 23 * 22 * 21 * 20 * 19)
= 0.02 (to 2 decimal places)

Because there is a 12/24 chance of selecting 1 red, then a 11/23 chance of selecting the next.

Then for the next part, since the probability is symmetrical

P = (1 - 0.02)/2
= 0.49

Where did I go wrong?

Also, do you think I will get 1/2 for part ii since my method is right?

 

[Mooju]

yeah you should get 1 /2 marks =] you went wrong with your part 1 bit, what you did there was SPECIFIED that the order had to be : R R R, Y Y Y and by doing that you made ORDER very important. Thats why you needed to go

( 12C3 x 12C3 )/24C6

BECAUSE what that statement says is I choose 3 red from 12 and choose 3 yellow from 12 ---> the favourable outcomes and then divide it by the number of outcomes which is from 24 marbles i choose 6. If you put it into context it makes much more sense.

From there you use your method for part 2

 

[joshuajspence]

yeah you should get 1 /2 marks =] you went wrong with your part 1 bit, what you did there was SPECIFIED that the order had to be : R R R, Y Y Y and by doing that you made ORDER very important. Thats why you needed to go

( 12C3 x 12C3 )/24C6

BECAUSE what that statement says is I choose 3 red from 12 and choose 3 yellow from 12 ---> the favourable outcomes and then divide it by the number of outcomes which is from 24 marbles i choose 6. If you put it into context it makes much more sense.

From there you use your method for part 2

omg of course.... argh i hate probability... thanks Mooju.

 

[Mooju]

Not a problem ^^ im doing one for 3u now T_T its soo hard im so stuck T_T XDD

 

[bikapika]

Initially, that's what I thought of, but when you look at the question, order isn't important. Think about it this way, when you grab 3 marbles right? of any colour, is the order you pick them out important if you're concerned with what the final result is, it's exactly like the question in the exam. You've got 12 marbles, 6 red and 6 yellow, and the question is concerning itself with 3 reds and 3 yellows, not to exactly have 2 reds, a yellow a red and then the other 2 yellows if you catch my meaning. It is concerned with a GROUP of 3 reds and 3 yellows not the ORDER of 3 reds and 3 yellows. That's why I still agree with Terry's method in doing it.

Plus also what my teacher said at one point. Always ask yourself if order is important. Generally that'll tell you whether you're dealing with perms or combs. In this case about puicking out and worried about just have how many balls of what in the end result, I concluded it was combinations.

If THAT didn't make sense I'm sorry

XD

Really lousy when it comes to explaining probability and stuff --> it gets confusing T_T

sorry i still dont get it actually i put down terry's method at the beginning and then crossed it out as i didnt feel right about it

I think you and Terry were trying to say:

Favourable outcomes:
select 3 yellow and 3 red from 12 yellow and 13 red
i.e. 12 x 11 x 10 x 12 x 11 x 10 is one outcome, but since all you need is 3 red +3 yellow as an OUTCOME, not the ways to get this outcome, then another way such as 12 x 12 x11 x10 x 11 x 10 is still producing the same outcome. as a result 12C3 x 12C3 makes perfect sense

Sample Space:
all you need is to find out the possible outcomes by selecting 6 out of 24. so 24C6 makes sense.


My worry is that, the two methods are not producing the same answers. Hence one of the method must be wrong, and i cant see where the less efficient but more prudential one went wrong.

in addition, although ORDER is not a concern in this question, however, different orders DO produce additional ways to get the desired outcome. A 12C3 x 12C3 does not capture all possible ways to get the outcome that contains 3 red and 3 yellow.

Furthermore, the whole purpose of using 12C3 x 12C3 to capture ways of producing the desired outcome indcating the fact that the number of ways is important. if not then why dont we just go:

Favourable outcome = 1
Sample space:
0 yellow 6 red, 1 yellow 5 red, 2 yellow 4 red, 3 yellow 3 red, 4 yellow 2 red, 5 yellow 1 red, 6 yellow 0 red, a total of 7 different outcomes

so 1/7=0.14 for part (i) ? We dont do it this way because we know that number of ways that the favourable outcome/possible outcome can occur DOES matter, which makes the order matter.

am i making you guys more confusing?

 

[bikapika]

Not a problem ^^ im doing one for 3u now T_T its soo hard im so stuck T_T XDD

kidding me lol? 3u is like a joke to me and u can get 30+marks more than me in 4u

 

[Mooju]

kidding me lol? 3u is like a joke to me and u can get 30+marks more than me in 4u

ROFL it was a question 7 b) .... so its almost like targeting the 4u guys but in the end i solved it i was so happy when i figured it out. I hate it when they put probability like as the last question of the 3u paper. It can get really tough, lucky for me when I time myself, i finish q 1-6 in an hour and can spend an hour or so on question 7 =]

sorry i still dont get it actually i put down terry's method at the beginning and then crossed it out as i didnt feel right about it

I think you and Terry were trying to say:

Favourable outcomes:
select 3 yellow and 3 red from 12 yellow and 13 red
i.e. 12 x 11 x 10 x 12 x 11 x 10 is one outcome, but since all you need is 3 red +3 yellow as an OUTCOME, not the ways to get this outcome, then another way such as 12 x 12 x11 x10 x 11 x 10 is still producing the same outcome. as a result 12C3 x 12C3 makes perfect sense

Sample Space:
all you need is to find out the possible outcomes by selecting 6 out of 24. so 24C6 makes sense.


My worry is that, the two methods are not producing the same answers. Hence one of the method must be wrong, and i cant see where the less efficient but more prudential one went wrong.

in addition, although ORDER is not a concern in this question, however, different orders DO produce additional ways to get the desired outcome. A 12C3 x 12C3 does not capture all possible ways to get the outcome that contains 3 red and 3 yellow.

Furthermore, the whole purpose of using 12C3 x 12C3 to capture ways of producing the desired outcome indcating the fact that the number of ways is important. if not then why dont we just go:

Favourable outcome = 1
Sample space:
0 yellow 6 red, 1 yellow 5 red, 2 yellow 4 red, 3 yellow 3 red, 4 yellow 2 red, 5 yellow 1 red, 6 yellow 0 red, a total of 7 different outcomes

so 1/7=0.14 for part (i) ? We dont do it this way because we know that number of ways that the favourable outcome/possible outcome can occur DOES matter, which makes the order matter.

am i making you guys more confusing?

yeah 12C3 x 12C3 is saying the number of ways you can choose this ofcourse, but what about 12P3 x 12P3? The difference between 12C3 and 12P3 is that 12P3 considers the ORDER that you place them in.

What I mean is ok so lets say 12C3 is im gonna pick any of these 3 balls, and I could pick this ball A, ball B, then ball C but THAT is equal to also picking up ball B, ball A, then ball C yeah? But then in 12C3 you could also pick ball D, ball A, ball C. See how it's not realy number of ways of picking up the SAME balls? It's rather the number of ways of picking COMBINATIONS or balls. Where as 12P3 is saying that picking up ball A, ball B, then ball C is different from picking up ball B, ball A, then ball C, but in the context of the question, what difference is picking up ball A, ball B, then ball C from picking up ball B, ball A, then ball C?

There is no difference you know?

What you said about the 1/7 is also incorrect. Because what you have FAILED to consider also is that there are 24 balls. Sure you said 1R 6Y and so on, but at the same time, what happened to the other 18 balls? Did you take them into consideration? So that explanation becomes false, it isn't actually as you say it is. Because remember you think about 24 red balls: they all look identical therefore they are the SAME but, to pick out 6 red balls you could pick out:

R1,R2,R3,R4,R5,R6
OR
R2,R3,R4,R5,R6,R7 and so on

It'd still be 6 RED balls, but you see how the two combinations are different? You've picked up 6 DIFFERENT red balls and that is why you use 12C3 or 24C6. It doesnt matter which order the red balls in, but what red balls you do pick up.

 

[bikapika]

ROFL it was a question 7 b) .... so its almost like targeting the 4u guys but in the end i solved it i was so happy when i figured it out. I hate it when they put probability like as the last question of the 3u paper. It can get really tough, lucky for me when I time myself, i finish q 1-6 in an hour and can spend an hour or so on question 7 =]

Good luck lol, hope u get full mark

The difference between 12C3 and 12P3 is that 12P3 considers the ORDER that you place them in........Did you take them into consideration?

Yes I did!! Please look up on the top of the thread im sure i always put down 12C3 X 12C3 .etc.

the "order" element was imposed through multiplying the result of 12C3 x 12 C3 by 6P6 , and then divided by 3!3! to take consideration of the fact that yellow balls are identical and red balls are identical.

i diIt doesnt matter which order the red balls in, but what red balls you do pick up.

In my head, its still It doesnt matter which order the red balls in, but how many ways can u pick up the desired number of red balls.

 

[Mooju]

Ohhhhh sorry for misunderstanding you, but when you said 6P6 has 3! x 3! considered, 6P6 / 3!3! = 20 =S..... there's something you're not taking into consideration.... no i know why, its because what you said ultimately was 6P6 = 6! right (mathematically) so therefore you've said 6!/3! times whatever.

Hold that thought, then go to a problem that says how many arrangements you can make with 6 letters where 3 are the same and the other 3 are the same.

No of arrangements = 6!/3!3! ---> we consider ARRANGEMENTS / PERMUTATIONS here yeah?

i mean i understand what you were thinking cause I got stuck there as well.. confused at what was different but I think that's it. You still without realising considered arrangements ---> I THINK!!!!! >.<

Oh and thanks for the good luck!!! I wish you good luck too =DDD tomorrow it is!! >.<

 

[bikapika]

Hold that thought, then go to a problem that says how many arrangements you can make with 6 letters where 3 are the same and the other 3 are the same.

No of arrangements = 6!/3!3! ---> we consider ARRANGEMENTS / PERMUTATIONS here yeah?

YEP YEP YEP thats my idea of multiplying 12C3 X 12C3 with 6!/(3!3!) in order to get the total number of outcomes with 3 yellows and 3 reds

lol i sent Terry Lee an email about this question. Hope he replies although seems like hes a busy person

 

[Mooju]

]4) X 2 / (5) = 0.64

Question, why'd you multiply by 2?
[/FONT]

 

[bikapika]

because you will get repeated number of the same outcomes from
3 yellow + 3 red, and
3 red + 3 yellow,
while they are mutually exclusive results. Selecting exactly 3 red balls fulfills the questions, while selecting exact 3 YELLOW balls also fulfills the question

 

[Mooju]

but, if you choose 3 red balls then you automatically choose 3 yellow balls for the favourable outcomes.... still confused as to why you multiply by two because then arent you just doubling up one what isn't necessary?

 

[bikapika]

OMG thank you somuch mooju for taking my problem seriously T_T
most ppl with 110+ would just ignore a little 70ish kid like me T_T^99

 

[bikapika]

but, if you choose 3 red balls then you automatically choose 3 yellow balls for the favourable outcomes.... still confused as to why you multiply by two because then arent you just doubling up one what isn't necessary?

just to make it extremely somple, for example, if you have 4 balls, 2 yellow (Ya, Yb), 2 red(Ra, Rb), and then mix then in box just like this question.

now, you are asked to find out the probability of selecting 1 yellow and 1 red.

the favourable outcome is given by (alphabe arranged in the order of selection)

Ya Ra,
Ya, Rb
Yb, Ra,
Yb, Rb

and

Ra, Ya
Ra, Yb
Rb, Ya
Rb, Yb

I.e. a total of 8 ways. which equals to 2C1 x 2C1 x 2!........

ooops, proof myself wrong lol

omg im so funny

you are correct we dont need to times by 2, ill make the correction

 

[Mooju]

ROFLMAO AHAHAHAH man when there's a problem and someone says something logical I always get interested, when it comes to maths man I HAVE to solve the problem no matter what. I'm seriously just a stubborn bastard when it comes to tricky problems XDD

 

[Mooju]

ahahah from there you would've gotten it correct then! if you didnt multiply by two then you would've gotten the right answer =] so in fact you were SO CLOSE , just that 2 threw it off =[

and that last pose i made was the the one about me taking the question seriously xDDD i didnt see you wrote the explanation and then proved yourself wrong so i wasnt actually laughing that you were wrong XDDDDDDDDD

but WOOH at least we got that clear and covered =]

EDIT- hangon a sec... no what you've written there IS 8(2C1 x 2C1 x 2) different ways.. but it should only be 4 different ways....

i think i know where the problem lies...

a x b is the same as b x a --> i think that's where you are doubling up where you shouldn't be. get it?

I think with the intially post you made, the number (5) you might have missed a multiple of 2 because you considered it for (4)?

EDIT x 2 - dude i think i know where the problem lies

Ways for 0 red 6 yellow: 12C6=924.......................................... ........(1)
Ways for 1 red 5 yellow : 12C1 X 12C5 X 6P6/(5!) = 57024..............(2)
since you can withdraw the 6 marbles in any order (6P6 occurs) but some yellows are repeated ( 5! Occurs)
Ways for 2 red 4 yellow: 12C2 X 12C4 X 6P6/(2!4!) = 490050………….(3)
Ways for 3 red 3 yellow: 12C3 X 12C3 X 6P6/(3!3!) = 968000………….(4)

Now [(1) + (2) + (4)] X 2 + (3) = 2063996…….(5)
in order to take account of 0 yellow 6 red, 1 yellow 5 red ……..

you need to multiply (3) by 2 as you said, I was wrong because what you've done is you've multiplied EVERYTHING by 2 because like you said for your way, you need to consider both yellow and red's for all the number of outcomes.

but for the favourable, you dont multiply by 2... and for that.. im still trying to solve the answer too... xDDD ill come back to it tomorrow right now I gotta do some 3 u papers XDD good luck!!!!!