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Bunuel
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Very intelligent question where we attack what the author doesn’t want and subtract from the sample set.
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Why the probability of having a double 4 isn't (5/6)*(5/6)?
I understand it logically but not mathematically.
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Nasco
Why the probability of having a double 4 isn't (5/6)*(5/6)?
I understand it logically but not mathematically.

Probability of having first dice 4 is 1/6
Probability of having second dice 4 is 1/6

Then we have to use AND probability...
Probability of having two dice 4 is: 1/6*1/6
Therefore answer is 1/36.

Now probability of having not having double 4 is: 1-1/36= 35/36

Therefore answer.. Atleast one double 4= 1-(probably of having no double 4)

Thus: 1-35/36
For n times... 1- (35/36)^n

D

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Nasco
Why the probability of having a double 4 isn't (5/6)*(5/6)?
I understand it logically but not mathematically.

I think you're asking why the probability of NOT having a double 4 isn't (5/6)*(5/6).

(5/6)(5/6) represents the probability that neither die shows a 4, meaning both dice must show numbers other than 4. However, not having a double 4 means any combination where at least one die is not a 4, so any combination except (4, 4) is allowed. Combinations like (1, 4) or (4, 3) are possible, which wouldn't be included in (5/6)(5/6).

Hope that makes it clearer!
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Bunuel
Nasco
Why the probability of having a double 4 isn't (5/6)*(5/6)?
I understand it logically but not mathematically.

I think you're asking why the probability of NOT having a double 4 isn't (5/6)*(5/6).

(5/6)(5/6) represents the probability that neither die shows a 4, meaning both dice must show numbers other than 4. However, not having a double 4 means any combination where at least one die is not a 4, so any combination except (4, 4) is allowed. Combinations like (1, 4) or (4, 3) are possible, which wouldn't be included in (5/6)(5/6).

Hope that makes it clearer!
Thank you for explaining
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