Recent content by bizboy1

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    Probability - Coin Toss - Find Formula

    Homework Statement Suppose a fair coin is tossed n times. Find simple formulae in terms of n and k for a) P(k-1 heads |k-1 or k heads) b) P(k heads | k-1 or k heads) Homework Equations P(k heads in n fair tosses)=binomial(n,k)/2^n (0<=k<=n) The Attempt at a Solution I'm stuck...
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    Real Analysis: Proving an*bn converges to ab

    Ya I agree with the above posters. This type of question is a triangle inequality question.
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    Stochastic modelling, poisson process

    Looks right to me. It could be an error in the back of the book.
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    Probability. You roll, I roll game.

    My confusion was the word play. Play means exactly what? 5 rolls total? That doesn't make sense. So it must be 5 complete games. Game includes the complete description.
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    Probability. You roll, I roll game.

    Answer: Probability of a win is 15/36. Now then use binomial(5,4)*(15/36)^4*(21/36)+binomial(5,5)*(15/36)^5
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    Probability. You roll, I roll game.

    I got it. The trick was applying the binomial theorem to the wins and not to the individual rolls.
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    Probability. You roll, I roll game.

    I think I figured it out. The problem with me and these questions is that I read them wrong/differently. Anyone else confused by the five games?
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    Probability. You roll, I roll game.

    Homework Statement You roll a die, and I roll a die. You win if the number showing on your die is strictly greater than the one on mine. If we play this game five times, what is the chance that you win at least four times? The answer is .1005 [b]2. Related Equations[\b] Binomial The Attempt...
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    Basic Probability Question—Rolling Die

    There's two ways of doing it. Short way: given that you rolled a number, the probability of getting that number on the second roll is 1/6. But since we didn't get that number, we have 5/6. Now, the probability of getting one of those two numbers is 2/6. So we multiply 5/6*2/6=10/36. The way...
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    Basic Probability Question—Rolling Die

    Actually I just figured it out. It's similar to the birthday problem. Just create trees for multiplication rule for n events and you will get it. p4=(5/6)*(4/6)*(3/6) p5=(5/6)*(4/6)*(3/6)*(4/6) p6=(5/6)*(4/6)*(3/6)*(2/6)*(5/6) p7=(5/6)*(4/6)*(3/6)*(2/6)(1/6)*(6/6)
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    Basic Probability Question—Rolling Die

    Basic Probability Question—Rolling Die!!! Homework Statement Suppose you roll a fair six-sided die repeatedly until the first time you roll a number that you have rolled before. A) for each r=1,2,... calculate the probability pr that you roll exactly r times. Homework Equations...
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