Probability question (1 Viewer)
- Thread starter apollo1
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From i) we know that:
P(odd one out)=
Now for ii) we are now playing N games.
We want to find the probability of getting at least one odd out in N games.
P(at least one odd out in N games) = 1-P(no odd outs in N games)
P(no odd one outs in one game) =
Therefore,
P(no odd one outs in N games) =^{N})
Hence,
P(at least one odd out in N games) =^{N})
part iii) and iv) are coming soon
P(odd one out)=
Now for ii) we are now playing N games.
We want to find the probability of getting at least one odd out in N games.
P(at least one odd out in N games) = 1-P(no odd outs in N games)
P(no odd one outs in one game) =
Therefore,
P(no odd one outs in N games) =
Hence,
P(at least one odd out in N games) =
part iii) and iv) are coming soon
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For iii) we want the first odd one out to occur on the Nth game.
P(odd one out occuring on Nth go) = P(no odd one out)xP(no odd one out)x...P(no odd one out)xP(odd one out), where there are N-1 P(no odd out out).
So in mathematical terms:
Let E=odd one out occuring on Nth go
Therefore:
P(E)=*(1-\frac{n}{2^{n-1}})*...*(1-\frac{n}{2^{n-1}})*(\frac{n}{2^{n-1}}))
Remember there are N-1 terms for)
,
So,
P(E)=^{N-1}*(\frac{n}{2^{n-1}}))
part iv) coming soon
P(odd one out occuring on Nth go) = P(no odd one out)xP(no odd one out)x...P(no odd one out)xP(odd one out), where there are N-1 P(no odd out out).
So in mathematical terms:
Let E=odd one out occuring on Nth go
Therefore:
P(E)=
Remember there are N-1 terms for
So,
P(E)=
part iv) coming soon
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yeh i forgot when you times them N times they are the same so it is to the power of N lolthe answers say part (ii) is:
^N)
For part iv) we want:
P(second odd one out occurs on Nth game)
Let A = second odd one out occurs on Nth game
Now this means the first odd one out can occur either on the first go up to the (N-1)th go. So:
P(A)=*(1-\frac{n}{2^{n-1}})*(1-\frac{n}{2^{n-1}})*...*(1-\frac{n}{2^{n-1}})*(\frac{n}{2^{n-1}}))
but dont forget that the first odd one out can occur in any of the first N-1 games, so this means there will be N-2 no odd one outs and hence:
P(A)=(\frac{n}{2^{n-1}})^{2}*(1-\frac{n}{2^{n-1}})^{N-2})
P(second odd one out occurs on Nth game)
Let A = second odd one out occurs on Nth game
Now this means the first odd one out can occur either on the first go up to the (N-1)th go. So:
P(A)=
but dont forget that the first odd one out can occur in any of the first N-1 games, so this means there will be N-2 no odd one outs and hence:
P(A)=
i know it's correct, however my aim was for you to understand it all lol, hope it helped
And yet again another great question of yours lol
apollo1
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yeah it helped heaps. i just said it was correct because its probability and u never quite no the answer with it.
lol well in this case i knew it had to be right, this was one of those prob questions if you lay it out right you will 100% know you are right