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# S1 Jun '09 Help Please? Watch

1. 9) Repeated independent trials of a certain experiment are carried out. On each trial the probability of success is 0.12.
(i) Find the smallest value of n such that the probability of at least one success in n trials is more than 0.95. [3]
(ii) Find the probability that the 3rd success occurs on the 7th trial. [5]

I have no idea where to even start on this one.
I looked at the mark scheme and it doesn't really give me a clue as to what it's asking for.

A little shove in the right direction would be much appreciated!
2. (Original post by Dusky Mauve)
9) Repeated independent trials of a certain experiment are carried out. On each trial the probability of success is 0.12.
(i) Find the smallest value of n such that the probability of at least one success in n trials is more than 0.95. [3]
(ii) Find the probability that the 3rd success occurs on the 7th trial. [5]

I have no idea where to even start on this one.
I looked at the mark scheme and it doesn't really give me a clue as to what it's asking for.

A little shove in the right direction would be much appreciated!
P(At least one success) equals 1 minus P(all fail).

So what's the probability of n failures, and you want this to be less than 1-0.95
3. (Original post by ghostwalker)
P(At least one success) equals 1 minus P(all fail).

So what's the probability of n failures, and you want this to be less than 1-0.95
I don't really understand ?? Sorry
The first mark in the mark scheme is awarded for (1-0.12)^n and I can't see why.
4. (Original post by Dusky Mauve)
I don't really understand ?? Sorry
The first mark in the mark scheme is awarded for (1-0.12)^n and I can't see why.
The probability of a success is 0.12, so the probability of a fail is 1-0.12.

The probability of n failures in a row is thus (1-0.12)^n
5. (Original post by ghostwalker)
The probability of a success is 0.12, so the probability of a fail is 1-0.12.

The probability of n failures in a row is thus (1-0.12)^n
Thanks very much, got the first part now. Still trying to get my head around the fact that P(X>=1)=1-P(1-p)
Had no trouble with part two surprisingly. Thanks!
6. (Original post by Dusky Mauve)
Thanks very much, got the first part now. Still trying to get my head around the fact that P(X>=1)=1-P(1-p)
Had no trouble with part two surprisingly. Thanks!
Good that you got the second part - which is tricky in its own right.

Don't follow your statement for the first part. Perhaps this will help.

P(X>=1) = 1-P(X=0)

This is because there are either no successes, or there is one or more successes, and because that covers all possibilities and the events are mutually exclusive (you can't get 0, and be >=1 at the same time),
so P(X>=1)+P(X=0) = 1, and hence....

Considering just a single trial we have P(fail) = 1- P(success)
7. (Original post by ghostwalker)
Good that you got the second part - which is tricky in its own right.

Don't follow your statement for the first part. Perhaps this will help.

P(X>=1) = 1-P(X=0)

This is because there are either no successes, or there is one or more successes, and because that covers all possibilities and the events are mutually exclusive (you can't get 0, and be >=1 at the same time),
so P(X>=1)+P(X=0) = 1, and hence....

Considering just a single trial we have P(fail) = 1- P(success)
I think that makes a bit more sense, thanks!
My brain just isn't made for stats

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