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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...

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...

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...

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, yes there are 4 different ways, but each way has 2 different ways/approaches to get, as a result...

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...

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

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

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

Good luck lol, hope u get full mark :) 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...

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

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...

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...

lol cool after reading terry lee's paper i would get 84/120, and then 84 X 90% for chances of mistakes that i didnt take account, lol 75.6/120!!! from some prediction this could still make it to low e4 (please be 90 exactly)

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...

yeah but dont u have to take account into the order of withdrawing balls?

u will love economics in year12 :) i thought of it while i was doing naughty stuff with naughy ppl

just a funny question in relation to the 5 sector model: assume that 1.) t + s = g + i 2.) the economy has a floating currency in this case, theorically if m>x, the economy is supose to contract while if m<x, the economy will expand. now, if u take into account of the balance of payments...

omg i feel so sorry for u. dont worry things will work out. usually they look at the overall performance of your cohorts and then determine an appropriate mark for u. im sorry but u dont get to resit the exam.

hey tracy good to see u again (remember esl?) so how did u go with the other exams? feelin aright?

thats the same as me lol. well almost, since i did about 1/3 of textbook questions and past papers since 1998 prior to the hsc, omg and ur raw mark is 30~40 more than mine and u studied less lol i was so confident of getting 90+ at the beginning and then due to poor time management i skipped a...