91Ó°ÊÓ

(a) The sector $X$ to the quantity of dice they means the amount one.

(b) But there are six dice, factor $X$ also can have variables of $0,1,2,3,4,5,6$.

(c) (c) Characteristic $X$futures a multivariate model with six trials and a $\frac{1}{6}$rate of occurrence. But we've got six dice, there are six trials. So there are six values on the dice or the possibilities of getting much loved is $\frac{1}{6}$, the possibilities of occurance is $1$.

(d) We all know that perhaps the calculated return of a random vector with $n$trials and $p$completion likelihood is that$6×\frac{1}{6}=1$. As a conclusion, variable $X$to need a results of localid="1649852478773" $n×P$.

(e) Our variable $X$takes on the worth $6$when all six dice land on one. As a result, the desired probability is provided by

$P(X=6)=\left(\begin{array}{l}6\\ 6\end{array}\right){\left(\frac{1}{6}\right)}^6=0.00002$

(f) The likelihood of three dice displaying the quantity one is:

$P(X=3)=\left(\begin{array}{l}6\\ 3\end{array}\right){\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^3=0.05$

The prospect of $4$dice displaying the amount one is as follows:

$P(X=4)=\left(\begin{array}{l}6\\ 4\end{array}\right){\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^2=0.0075$

We may now conclude that three dice will possibly display a $1$.