Respuesta :
The probability that every even number appears at least once before the first occurrence of an odd number is [tex]\frac{1}{20}[/tex]
Take [tex]A_{k}[/tex] - the event that odd number appeared on the k-th throw, B - every even number appeared at least once.
Let’s find P(B|[tex]A_{k}[/tex]) . Note that this probability is 0 for k∈{1,2,3} as you need k≥4 to place all distinct even numbers before the odd one.
Now for k≥4 we need to use the formula of inclusions and exclusions: P(B¯|[tex]A_{k}[/tex])=P(C2+C4+C6|[tex]A_{k}[/tex])=
=P(C2|[tex]A_{k}[/tex])+P(C4|[tex]A_{k}[/tex])+P(C6|[tex]A_{k}[/tex])−P(C2C4|[tex]A_{k}[/tex])−P(C2C6|[tex]A_{k}[/tex])−P(C4C6|[tex]A_{k}[/tex]) where Ci is the event that the dice i is missing.
These probabilities are:
P(Ci/[tex]A_{k}[/tex])=[tex](\frac{2}{3} )^{k-1}[/tex]
P(CiCj|[tex]A_{k}[/tex])=[tex](\frac{1}{3} )^{k-1}[/tex]
So
P(B|[tex]A_{k}[/tex])=1–P(B¯|[tex]A_{k}[/tex])=1−3⋅[tex](\frac{2}{3} )^{k-1}[/tex]+3[tex](\frac{2}{3} )^{k-1}[/tex].= 1 − [tex]\frac{9}{2}[/tex].[tex](\frac{2}{3} )^{k}[/tex]+9[tex](\frac{1}{3} )^{k}[/tex]
Now as P([tex]A_{k}[/tex])= [tex]\frac{1}{2^{k} }[/tex]we conclude that
P(B) = ∞∑k=4P(B/[tex]A_{k}[/tex])P([tex]A_{k}[/tex])= ∞∑k=4([tex](\frac{1}{2} )^{k}[/tex]−[tex]\frac{9}{2}[/tex][tex](\frac{1}{3} )^{k}[/tex]+9.[tex](\frac{1}{6} )^{k}[/tex])=
=[tex]\frac{\frac{1}{16} }{\frac{1}{2} }[/tex]−[tex]\frac{\frac{1}{18} }{\frac{2}{3} }[/tex]+[tex]\frac{\frac{1}{144} }{\frac{5}{6} }[/tex] = [tex]\frac{1}{8}[/tex] -[tex]\frac{1}{12}[/tex] + [tex]\frac{1}{120}[/tex] =[tex]\frac{6}{120}[/tex] = [tex]\frac{1}{20}[/tex]
the probability that every even number appears at least once before the first occurrence of an odd number is [tex]\frac{1}{20}[/tex]
To learn more about even numbers:
https://brainly.com/question/2289438
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