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

[bikapika]

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)?

you can feel free to laugh as loud as u want to lol ppl at my school laught at me for all kinds of reasons and i ready get used to it and im proud of being funny

anyway get back to serious staff,

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?

yes there are 4 different ways, but each way has 2 different ways/approaches to get, as a result, if we want to find the total number of ways, it should be 4 x 2 yea?

oh btw terry lee replied my email omg hes such a nice guy, and then he started to mention what we've discussed earlier, im trying to keep approaching him for further explaination

 

[bikapika]

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!!!!!

No no no you are correct!! since my approach for (3) was:
12C3 x 12C3 x 6! / (3!3!), where this the bold part should have taken account into all the possible ways already by using permutation of 6 to arrange all the 6 balls.

the logical reason why i multiply (1), (2) ... by 2 was because:

1 yellow + 5 red and 1 red +5 yellow

are completely different selections but they are symmetric

while 3 yellow 3 red is the same as 3 red and 3 yellow, so we shouldn't multiply by 2. Its like for binomial probability with even power, the middle coefficient is alone without symmetrical partner

as proved in my proof that proves myself wrong (look at my poor language skills ):

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

The complete working out should be:

2C1 x 2C1 X 2! /(1!1!) = 8 ways, no need to multiply by 2.

however, if you are asked to select 2 yellow balls out of a box of 2 yellow + 2 red,

then you will get:

Ya Yb

or

Yb Ya,

i.e. 2 ways = 2C2 x 2! /2! = 2

And then you multiply by 2 to include selection of 2 red balls out of the box,

Ra Rb
Rb Ra

 

[bikapika]

OMG why dont we just use binomial probability to find it!!!?????

Let (y+r)^6=y6 + 6C1 y^5 r + 6C2 y^4 r^2 +6C3 y^3 r^3 + 6C4 y^2 r^4 + 6C5 y1 r^5 + r^6

Number of ways to select 3 yellow and 3 r is obtained from the term:

6C3 x y^3 r^3

As you are selecting 3 yellows and 3 reds out of 12 yellows and 12 reds, so:

6C3 x 12C3 x 12 C3 = 12C3 x 12C3 x 6! / (3!3!) = 968000 same as my method

 

[Mooju]

i dont know if you can use binomial =/, because for binomial to work you need to have repeating elements but in a basket full of 24 balls, once you take one out you dont put another one in to replace it you know what i mean?

 

[bikapika]

Yes i know what you mean, however i dont think whether the elements are repeating or not will matter in this case.

This is happening as:

see for y^6 or 6C1 y^5 r or other binomial terms, they are completely different and exclusive trials, each one of them will be considered individually.

Since the entire purpose of binomial is to gain a clear image about how many ways can 6 balls be made up with yellow or red, regardless of how many yellows or reds you can select from.

For example, for 6C3y^3 r^3, binomial theorem tells us that "there are 6C3 ways you can select 6 balls with 3 "y"s and 3 "r"s THATS IT.

And then we start to calculate how many 3 "y"s can you select from 12 "y"s and so on....


Seriously binomial is fun although i dont really get the whole image behind why (A+B)^N= sum (r=0 to r=n) nCr A^(n-r)B^r - -my textbook doesnt prove this formula

 

[Mooju]

i suppose i see what you're saying... still not really convinced but nywho, this argument could go on FOREVER xDDDD has terry lee replied yet?

 

[bikapika]

he replied my first email but thats it. he mentioned the exact same argument you gave to me at your first post reply.

Hope he replies the second email when he has time.

 

[Mooju]

yeahh i still support my reasoning, but i guess i understand yours a bit, im just relaly glad that this whole topic didnt confuse me for 3u today XDDD

 

[bikapika]

lol so how did u go, really good?
i think i could get about 75

 

[Mooju]

did awesome mate!!! came out thinking 100% looked at terry lee's answers and now 83/84!!!! loving it!! =DD

and congrats on the 75! that's quite an achievement too ^^

 

[bikapika]

did awesome mate!!! came out thinking 100% looked at terry lee's answers and now 83/84!!!! loving it!! =DD

and congrats on the 75! that's quite an achievement too ^^

after checking with terry lee's answer, i made a few mistakes so 70 -0- still good enough for e4, i think?

omg im so proud of u. Good luck for your future mate hope u become a successful doctor/lawyer/engineer/actuary and other 99.9+ HSC professions

 

[Mooju]

should be, from reading the extension i threads, apparently the exam was pretty hard (albeit not as hard as last years, but still pretty hard) so you should still get e4 for that hopefully, i wonder how e4 translates out of 100 thought , probably double the mark.

and funny that i dont wanna be any 99.9 hsc profession xDDDDD just gonna do maths in uni and become a code breaking cryptography da vinci style ROFLMAO AHHAHAHAHA xDD

 

[bikapika]

should be, from reading the extension i threads, apparently the exam was pretty hard (albeit not as hard as last years, but still pretty hard) so you should still get e4 for that hopefully, i wonder how e4 translates out of 100 thought , probably double the mark.

and funny that i dont wanna be any 99.9 hsc profession xDDDDD just gonna do maths in uni and become a code breaking cryptography da vinci style ROFLMAO AHHAHAHAHA xDD

Well my idea of a good profession is about doing what you enjoy to do and feel happy about going to work everyday as long as you are happy, then you are in a good job

Wish you all the best lol im still waiting for terry lee's reply although i dont think hes goin to reply again

 

[Mooju]

ahahah that's exactly what I think too ^^ good luck with all your exams too!!! still so curious about what terry lee says about it too T_T i wanna know!!!! *whines*

 

[bikapika]

ahahah that's exactly what I think too ^^ good luck with all your exams too!!! still so curious about what terry lee says about it too T_T i wanna know!!!! *whines*

he doesnt reply me no more T_T