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1. (Original post by Glavien)
Ohhh, that took me a while too understand. Thanks for that.
No problem.
2. (Original post by Glavien)
Ohhh, sorry about that, I was doing P(1 and 1) P(3 and 3) P(5 and 5) twice by mistake. Thank you!!
May I just add in (it's two weeks old but..) you could just find the probability that it is odd (so add P(x=1) + p(x=3) + p(x=5)) and then square it and get the same result. So you get (0.62)^2 = 0.3844. It's a bit quicker than finding all the potential ones!
3. (Original post by iMacJack)
May I just add in (it's two weeks old but..) you could just find the probability that it is odd (so add P(x=1) + p(x=3) + p(x=5)) and then square it and get the same result. So you get (0.62)^2 = 0.3844. It's a bit quicker than finding all the potential ones!
Yes, he's aware of that.
4. (Original post by Zacken)
Yes, he's aware of that.
5. (Original post by iMacJack)
6. (Original post by Zacken)
Yep I just saw! I still don't fully get it though, hence another response on the other thread
7. (Original post by iMacJack)
Yep I just saw! I still don't fully get it though, hence another response on the other thread
Ah, alright.
8. Hi, do we include the outlier in calculations?
9. (Original post by THESTRESS)
Hi, do we include the outlier in calculations?
No? Your post has no context and thus doesn't have a reasonable reply. Plus, you'd be better off posting this in another thread.

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