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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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    What have you come up with both questions? If you post your working here, we can see where you are going wrong and can guide you. Use the hints given to you in the question.. its a geometric progression
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    (Original post by y.u.mad.bro?)
    What have you come up with both questions? If you post your working here, we can see where you are going wrong and can guide you. Use the hints given to you in the question.. its a geometric progression
    I know that was a geometric progression theres a common ratio and I know the sum forumula but don't know how to get each part of the formula( a and r)

    With the second question, in the first part I was thinking 1/5*1/4*1/3... but that seems wrong. I know that the number of ways to order 5 numbers is 5! For the second part of the question I have no clue.

    This is also in the wrong forum so I posted it in the maths forum as well because I don't know how to delete this thread/ change the forum it is in
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    Let's take the basic step then.

    For Q1, you know that the chance is 10%. It is independent so it will always remain 10%. Therefore, your probability will be 0.1 for all chances. I don't know if you have yet studied geometric distribution but you know the probability now and you know that the question wants the success rate for the machine paying out for the first time on an odd numbered throw. Therefore, use these two and see what you can do. Try looking up geometric distribution if you haven't because that would be the method I would use to solve the question.

    For Q2, you are on the right track for the first part. For second part, how many different ways are there of you obtaining a number with 3 in the middle? Can you go on from there?
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    1)
    to win on the first throw - 0.1
    2- 0.9*0.1(lose on the first win on the second)
    3-0.9*0.9*0.1(lose 1 , lose 2 win third)
    4-0.1*(0.9)^3
    5- 0.1 *(0.9)^4
    the probability to win on odd throw is 0.1 + 0.1*(0.9)^2 + 0.1 * (0.9)^4 + ...
    geometric progression x1=0.9 and r=(0.9)^2
    to calculate this infinite sum you need to calculate the limit of the partial sums to infinity
    so you get lim ( 0.1(1-(0.9)^2n)/1- (0.9)^2
    = 0.1/0.19 which roughly equals 52%
    2) i) your approach is good
    ii) Think about what events have to occur so on your third pick you would get a 3 ( hint: you must have a 3 to pick from on your third pick )
 
 
 
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