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For part iv) we want: P(second odd one out occurs on Nth game) Let A = second odd one out occurs on Nth game Now this means the first odd one out can occur either on the first go up to the (N-1)th go. So: P(A)=...

yeh i forgot when you times them N times they are the same so it is to the power of N lol

For iii) we want the first odd one out to occur on the Nth game. P(odd one out occuring on Nth go) = P(no odd one out)xP(no odd one out)x...P(no odd one out)xP(odd one out), where there are N-1 P(no odd out out). So in mathematical terms: Let E=odd one out occuring on Nth go...

From i) we know that: P(odd one out)= \frac{n}{2^{n-1}} Now for ii) we are now playing N games. We want to find the probability of getting at least one odd out in N games. P(at least one odd out in N games) = 1-P(no odd outs in N games) P(no odd one outs in one game) =...

well if you want to put a number to it...i'd say around 7-8 hours a wk...1 hour per week day, 2-3 hours on sat, and leave sunday as a break cause you deserve it lol...i say you need a lot cause by the sounds of it 40 ppl doing mx2 means it will get very competitive and the other kids in the...

From i) we know that: [{\int_{0}^{a}}f(x)g(x)dx]^{2} \leq \int_{0}^{a}[f(x)]^{2}dx\int_{0}^{a}[g(x)]^{2}dx if we let: g(x)=a=1 into the above inequality we get: [{\int_{0}^{1}}f(x)dx]^{2} \leq \int_{0}^{1}[f(x)]^{2}dx\int_{0}^{1}1^{2}dx which simplifies to...

As a>o the original integral has to also be > 0 (due to the square) so that means a, b and c in the quadratic are greater than 0, and we form a positive definite quadratic, we know its positive definite as a and c will be greater than b

i would only focus on the questions you COULDNT get right, no reason to look at questions you know are easy and can do easily

There are in total 3n balls. For i) The probability of getting the same colour when 3 balls are drawn is as follows(noting there is no replacement): P(same\: \: colour)= 3[\frac{n}{3n}\frac{n-1}{3n-1}\frac{n-2}{3n-2}] Note it will be the same probability if it were red, blue or white so...

well with these possibilities...you work out how many ways you can have each: 3R 1Y 1G can be arranged in: \frac{5!}{3!} ways to count for the repetition of the 3 R's. 2R 1Y 2G can be arranged in: \frac{5!}{2!2!} ways to count for the repetition of the 2R's and G's Now you repeat this for...

Yeh the expansion of (1+x) to the n, if n is odd will always have two greatest coef's as i did say before. And for your above example, sub k=6 in and see if it is the same answer, then get back to me The reason for the equality part is there can be two greatest terms as i said before, so subbing...

From ii) [\sum_{k=1}^{n}(a_{k}^{2})]\geq \frac{1}{n}([\sum_{k=1}^{n}(a_{k})])^{2} now, let a_{k}=2k-1 sub this in: [\sum_{k=1}^{n}(2k-1)^{2}]\geq \frac{1}{n}([\sum_{k=1}^{n}(2k-1)])^{2} Expand each series out as follows: (2(1)-1)^{2}+(2(2)-1)^2+...+(2n-1)^2\geq...

And to actually answer this question: We want to find the greatest coef of the expansion, and the greatest coef is given by: \frac{T_{k+1}}{T_{k}}\geq 1 Now some tests actually make you derive the formula for the greatest coef, so ill do that here: First remember the general term T_{k+1} is...

What this is actually trying to work out is which term (k+1) is greater than every other term k, the reason for the \geq is cause there can sometimes be two greatest terms such in the expansion of (1+x)^{3}= 1 + 3x + 3x^{2} + x^{3} we see here that there are two greatest terms, T_{2}...

for the first one there really isnt anything better you can do..other methods would be to work out total number of combinations minus all the cases we dont want, but that is pretty long for this one too. For the second question there are 8 letters and 4 numbers...we want to pick two letters...

i hope this is a troll lol

you can also integrate it straight away realsing that cosx is the derivate of sinx using the reverse power rule for integrals: \int f'(x)[f(x)]^{n}dx= \frac{[f(x)]^{n+1}}{n+1}+ C which is in the cambridge 3u book..but this is more a 4u formula

by remainder thm, P(a) = R(x), where R(x) is your remainder, and the remainder is always at least one degree less than the divisor Edit: above polynomial if of form P(x) = A(x)Q(x) + R(x), where A(x) is your divisor, Q(x) is your quotient and R(x) is your remainder

let u= sinx, \frac{du}{dx}=cosx, du=cosxdx we now sub sinx for u and cosxdx for du as follows: \int cosxsin^{2}xdx=\int sin^{2}xcosxdx=\int u^{2}du which im sure you can finish

let u = sinx, then you can do it by substitution easy