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Let M=Outcome Bob has more Heads than Penny. P(M)+P(not M)=1 mutually exclusive and complementary (of course) But P(not M)=P(Bob has more Tails than Penny) -check this out assuming Bob has n+1 and Penny n coins. By symmetry P(M)=P(not M)=1/2 P(more heads)= P(more tails)

Bob tosses 11 coins, Penny tosses 10. (i) What is the probability that Bob has more heads than Penny. (ii) Generalize to n+1,n respectively.

Method 2 Let p= P(Bob wins) Then p= 1/4+1/2*1/4+1/2*1/4*p+1/4*p where 1/4=P(Bob wins first toss) 1/2*1/4=P(Bob wins on second toss after a 1 Head toss) 1/2*1/4=P(TT after a 1 Head toss) 1/4=P(TT on first toss) Solve for p to get 3/5. Yes...

___________________________________________________ Quote deyveed Whoa. Would a question like this be in the exam? It seems very hard and something that takes ages to learn. ___________________________________________________ The truth is "no holds barred" in these last questions, can't...

An example of how Penny can win : first toss HT, second toss TH - exactly one head happened each toss.

__________________________________________________ Quote from flyin' Since the probability that both (Bob and Penny) will win are equally likely in one attempt ... __________________________________________________ Just to clarify that Bob can win in just one attempt, but Penny requires...

ND : for a solution approach, check out 1999 HSC 7b.

Two players bet on the outcome of a toss of two coins. Bob bets that double heads will be tossed first. Penny bets that two consecutive single head will be tossed first (that is, exactly one head and one tail, Penny wins if this happens twice, one after the other). They keep tossing until one...

Nice work Archman. _________________________________________________ Quote Harimau: I feel so inferior everytime i look at someone solving all these hard questions.... _________________________________________________ Don't be discouraged, a question like this in the HSC (Q7 or 8) comes...

I wished!

__________________________________________________ What I have to do in this case, is to integrate dx from _0_ to x, rather than from H to x - and at the moment I can't quite figure out why - any thoughts? ___________________________________________________ Good students labour this point...

Announcement For most students in this forum, the interesting and lively on-going discussion of problems and solutions in these threads is sufficient extra-school mathematical study material. However, for some students who may need private, confidential one-on-one help, I am available online...

Use integral limits in solving mechanics rather than the more cumbersome 2-unit "integrate, add constant and check boundary conditions" way. See example in "trivial mechanics" thread.

Not really difficult (like a HSC Q7 polynomial question, all bark no bite), but adds depth to your polynomial knowledge. Let P(x) be the quadratic ax^2+bx+c. Suppose that P(x)=x has unequal roots. Show that the roots are also roots of P(P(x)=x. Find a quadratic equation for the other roots of...

Quote ezzy85______________________________________________ vdv/dx = -kv dv/dx = -k dv = -k dx I{1200--->800} v dv = I{0--->1400} -kx dx ________________________________________________ Just a correction on your integrals: I{1200--->800} dv = I{0--->1400} -k dx As well if you...

Thats right. Find k and use dv/dt. Very good. Wanted to find out whether students know how to use integral limits in solving mechanics this way rather than the more cumbersome 2-unit "integrate, add constant and check boundary conditions" way. ezzy85 passes the test!

A particle moves in a straight line subject only to a resistive force proportional to its speed. Its speed falls from 1200 m/s to 800 m/s over 1400 m. Find the time taken to the nearest 0.01 sec. Note :if writing integrals w/ limits, use notation I{a--->b}f(x)dx integral upper limit b...

or 1/(1+tan^.1(x)) + 1/(1+cot^.1(x)) =1/(1+tan^.1(x)) + tan^.1(x)/(1+tan^.1(x)) =1 since cot=1/tan. Quite useful to remember 1/(1+tan)+1/(1+cot)=1

freaking_out: 1/(1+tan^.1(x)) + 1/(1+cot^.1(x)) =1

A good problem, but demands some algebra using the textbook way ie. twice tangent means two pairs of double roots a,a,b,b. I would be interested in an "elegant" shortcut if you can remember it.