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A person throwing a dice and when even have come he got $2$ points and when odd have come he got $1$ points.What is the probability when he got exactly $6$ points?

edited
I make the case like this $P(6)=P(E)P(E)P(E)+P(E)P(E)P(O)P(O)+P(E)P(O)P(O)P(O)P(O)+P(O)P(O)P(O)P(O)P(O)P(O)$

for $P(E)=\frac{3}{6}=\frac{1}{2}$ and $P(O)=\frac{3}{6}=\frac{1}{2}$

So,$P(6)=\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}$

$P(6)=\frac{1}{2^{3}}+\frac{1}{2^{4}}+\frac{1}{2^{5}}+\frac{1}{2^{6}}$
$P(6)=\frac{14}{64}$

but answer is $\frac{43}{64}??$
$3$ consecutive even numbers = $(\frac{1}{2})^3$

2 Even and 2 Odd numbers can be arranged in $\frac{4!}{2! 2!} = 6$ = $6 \times (\frac{1}{2})^4$

similarly count for other cases like 4 odd 1 even, all odd ,etc

In your explanation i got $\frac{45}{64}?$

Thanks now I got the answer:)

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