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
#SPJ4
