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    What are the assumptions which need to be made for a binomial and geometic model to be valid?


    1) Sheena travels to work by car. She parks in her car parking space on 2 out of the 5 days on average. Let X be the number of days out of a 5 day working week on which she can park in her favourite space.

    A 5 day working week in which she can park in her space on more than 3 days is a 'good' week. Find the probability that out of 7 randomly chosen 5 day working weeks, fewer than 2 are good weeks.



    2) A turn in a game begins by throwing 2 unbiased dice. If at least one of the dice shows a 6 then the turn ends. If neither dice show a 6 one dice is thrown again and the turn ends. The number of 6s obtained at the end is denoted by S.

    0 1 2
    a b 1/36

    Find a and show that b is 85/216.



    3) Adill and Beth are playing a game. Adill throws a fair die 3 times. The number of 6s that he obtains is denoted by A. Beth throws a coin repeatedly. The number of throws up to and including the first throw on which the coin lands head upwards is denoted by B.

    Find p(A=B)




    4. On a fairground stall the price is worth £20. p(success) is 2/35. It is 50p a go and she keeps playing until she wins. I have found the E(X) but then how do i calculate the probability that vicky pays more than the prize is worth?




    If you could help me with any of these i would be greatful. Its just revision for me. Thanks
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    For a binomial distribution, you don't really make any assumptions. You can use it whenever an event has only 2 possible outcomes. For a geometric distribution, from what I gather you mean by that, the individual trials have to be independent of one another (eg. rolling 2 successive dice). Now as fior your questions:

    1) A week will either be good or not. i.e. we have 2 outcomes, and 7 trials. So we can use a binomial distribution here. The only problem will be finding the probability of a week being good. A good week is a week in which you have 3 days parking in a 5 week period. We're also given the probability that on a given day she finds a parking spot. So what's the probability that she finds a spot on 3 days in a 5 day week?

    2) First thing we need to do is find the probability that we don't get a 0 on our first or second roll. i.e. we need:
    p (no 6's on first roll of 2 dice)\capp(no 6s on 2nd roll of 1 dice).
    Remember, any 2 dice rolls are going to be independent of one another.

    3)This question's slightly longer, although the way to think of this is:
    Adam rolls a die 3 times. So he'll either got 0,1,2 or 3 6's. On the other hand Beth could get a head on her 1st,2nd,3rd,......,100000th,.... toss.So for which of these numbers can both A and B be achieved? i.e. whats \begin{Bmatrix} 0,&1,&2,&3\end{Bmatrix}\cap \begin{Bmatrix}1,&2,&3,&...\end {Bmatrix}. Now for each of these what's the probability that Adam get that many 6's and what's the probability that Beth takes that many turns to get a heads? Now what's the probability that for each of these, Adam gets a heads AND that Beth takes that many turns?...from there all you need to do is add up your individual probabilities and you'll have your answer.

    4) Firstly you have to realise that on a given attempt, she'll either win it or lose it. If she wins, she stops (as far as we're concerned), and if she loses, she tries again. Now if she spends 50p a game she'll have spent £20 in 400 games. So essentially what it's asking is what's the probability that she loses in 400 consecutive games? i.e. let X be the number of games Vicky loses consecutively, work out p(X=400).

    I hope this helped.
 
 
 
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