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Anyone any good at "Odds"

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Anyone any good at "Odds"

Two unrelated odds one has a 4% probability of happening, the other has a 3% probability.
What is the probability of them both happening?
Instinct tells me to multiply, (.0012%) but I have a feeling it's the difference (1%)
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Re: Anyone any good at "Odds"

This is the probability of two independent events assuming that the second event is not dependent on the first occurring:
probability of (A and B) = probability of (A) x probability of (B)
4/100 x 3/100 = 12/10000 (3 in 2500) or, as you stated, a probability of 0.0012
Unless my maths has failed me I'm sure someone else will enlighten us both

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Re: Anyone any good at "Odds"

If totally unrelated it's as Mav states, i.e. 0.12%.
VileReynard
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Re: Anyone any good at "Odds"

And the probability of neither of them happening is 93.12%  Cheesy

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Re: Anyone any good at "Odds"

OK thanks ........does that make the probability of (either) one happening 7%?
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Re: Anyone any good at "Odds"

From memory I think the probability of either A or B {P(A or B)} = P(A) + P(B) - P(A and B)
=(4/100_7/100) -0.0012
=7/100- 0.0012
=0.0688
Hopefully someone will correct me if I'm wrong. Been a while...
Edited out erroneous decimal point.

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Re: Anyone any good at "Odds"

Looks correct to me I normally ignore (forget) the probability of A + B because it is very small  Cheesy
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VileReynard
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Re: Anyone any good at "Odds"

Quote from: Mav
From memory I think the probability of either A or B {P(A or B)} = P(A) + P(B) - P(A and B)
=(4/100_7/100) -0.0012
=7/100- 0.0012
=0.0688
Hopefully someone will correct me if I'm wrong. Been a while...

Correct! (because 0.0688 + 0.9312 makes unity!)

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Re: Anyone any good at "Odds"

I love it when a plan comes together Wink

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Re: Anyone any good at "Odds"

Thanks all, .........................drinks all round at the virtual bar Smiley
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