If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
General question. Are you allowed to start on the RHS for Trig Identities? I never have done that before.
Are you allowed to use MX2 techniques in MX1?
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Buy my books/notes cheaply here!
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Uni Course: Actuarial Studies and Statistics at MQ -- PM me if you have questions
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ATAR: 99.75
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If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Last edited by InteGrand; 23 Oct 2016 at 3:42 PM.
P(total) = P(H) + P(TH) + P(TTH) + P(TTTH) + ...
= p + (1-p)kp + ((1-p)^2)(k^2)p + ((1-p)^3)(k^3)p + ((1-p)^4))(k^4)p + ...
= p(1/(1-k(1-p)) (as k and (1-p) are less than one, the ratio is less than one and hence this is true)
= p/(1-k(1-p))
You can check if k=1 then the chance is 1 which makes sense, as eventually she will toss a heads no matter the probability if it is unchanging, and if k=0 then the probability is p, as she can only win if she tosses a head initially, as if she doesn't she can never win.
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