Probability question (1 Viewer)

apollo1

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Could someone help me with parts (ii), (iii) and (iv) of this question:
Screen shot 2011-10-13 at 11.02.40 PM.png
 

math man

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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) =

Hence,

P(at least one odd out in N games) =

part iii) and iv) are coming soon
 
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math man

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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)=

Remember there are N-1 terms for ,

So,

P(E)=

part iv) coming soon
 
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math man

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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)=

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)=
 

apollo1

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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
yeah it helped heaps. i just said it was correct because its probability and u never quite no the answer with it.
 

math man

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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
 

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