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    1.) A gambling machine is programmed to pay out on 10% of the times that it is played. Whether or not the machine pays out on any throw is independent of the outcome on any other throw.

    By using the formula for the sum of geometric progression(or otherwise) compute the probability that the machine pays out for the first time on an odd numbered throw.

    2.)A bag contains 5 balls which are number 1,2...,5. If balls are selected one at a time to form a five digit number, What is the probability that:

    i.The number formed will be 12345?
    ii. The middle number will be a 3?

    Im struggling to even start both of these questions. Help will be greatly appreciated.
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    For the first one, the probability that it pays out on the first throw is 0.1 The third go is 0.9*0.9*0.1. The 5th is 0.9*0.9*0.9*0.9*0.1

    Well you can spot the pattern. Pull out a factor of 0.1, first of all. 0.1(1+0.9*0.9+0.9*0.9*0.9*0.9+.. .) and so on. You can see that this is an arithmetic series with a formula ar^2(n-1)

    I'll let you solve it from there.

    For the second one, i) is easy enough. It's 1/5! <-- think about it? 1/5 * 1/4 * ... * 1/1

    ii) EDIT: over-complicated it and made a boo boo ... sorry!
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    (Original post by DrSebWilkes)
    For the first one, the probability that it pays out on the first throw is 0.1 The third go is 0.9*0.9*0.1. The 5th is 0.9*0.9*0.9*0.9*0.1

    Well you can spot the pattern. Pull out a factor of 0.1, first of all. 0.1(1+0.9*0.9+0.9*0.9*0.9*0.9+.. .) and so on. You can see that this is an arithmetic series with a formula ar^2(n-1)

    I'll let you solve it from there.

    For the second one, i) is easy enough. It's 1/5! <-- think about it? 1/5 * 1/4 * ... * 1/1

    ii) Well this is a bit trickier because order is really important. Effectively, we want the chance of not choosing it. You'll see why this is helpful. (4/5*3/4) is the chance of not choosing it, but it happen on the next. So we can do 1-(3/5)=0.4, which should be right because of the law that P(A)+P(A`)=1

    To be honest, I'm not totally sure about ii) so I rather you not necessarily follow it right away until someone confirms what I said.
    Thank you for your help. Just a question on the first part, could you do a sum to infinity (a/1-r). with a being 0.1 and r being (0.9)^2. This is because I'm not sure if it's asking the question in general or if there's 6 throws altogether
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    DrSebWilkes gave good answers for the first few. Part ii) is quite intuitive when you think about it. Clearly, any number is equally likely to be the middle number so the probability that the middle number is 3 must be 1/5
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    (Original post by Matt_Barber00)
    DrSebWilkes gave good answers for the first few. Part ii) is quite intuitive when you think about it. Clearly, any number is equally likely to be the middle number so the probability that the middle number is 3 must be 1/5
    Order is important though.

    Oh I just now actually, it would be 4/5*3/4*1/3

    Which is actually a 1/5 ... oh ... hahaha okay I take back what I said above.

    Still the ability to remove the probability of stuff not happening is still really handy in certain situations. Some people will say "Binomial distribution!!" but without knowledge of that, how would you answer a question like:

    "Each day there is a 20% chance of it raining? What is the chance it rains at least once within a 5 day period?"

    The easiet way to do it that effectively skips over the whole ordering issue is

    1-P(A`)=P(A). So, here we know that the chance of it not raining in 5 days is (4/5)^5

    so P(A)=1-(4/5)^5 = 0.67232
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    (Original post by Knowledgeis)
    Thank you for your help. Just a question on the first part, could you do a sum to infinity (a/1-r). with a being 0.1 and r being (0.9)^2. This is because I'm not sure if it's asking the question in general or if there's 6 throws altogether
    Indeed! That's exactly how to do it. As in yes, to infinity.
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    Thank you both for your help. I understand how ill approach similar questions next time
 
 
 
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