91影视

It has been given that Martian dice are regular tetrahedra with vertices labelled 1 to 4. Two such dice are tossed and the sum of the numbers showing is even.

Probability means the chances of any event to occur is called it probability.

a) The distribution is shown below so the sample space is given by

$\begin{array}{l}\text{S={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}\\ \text{(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)}}\end{array}$

Let x= even sum of numbers on dice which is given as mentioned below.

$x=\{2,4,6,8\}$

Find the probabilities.

$\begin{array}{l}P(x=2)=\frac{1}{16}\text{鈥夆赌夆赌�}\\ P(x=4)=\frac{3}{16}\\ P(x=6)=\frac{3}{16}\text{鈥夆赌夆赌�}\\ P(x=8)=\frac{1}{16}\end{array}$

Draw the distribution.

b) The number shown is even then for odd sum let x=0 so the expected value is given as mentioned below.

$\begin{array}{c}E\left(x\right)=渭\\ =鈭�xp\left(x\right)\\ =2\left(\frac{1}{16}\right)+4\left(\frac{3}{16}\right)+6\left(\frac{3}{16}\right)+8\left(\frac{1}{16}\right)\\ \text{=2.5}\end{array}$

Find the variance.

Find the standard deviation.

$\begin{array}{c}蟽=\sqrt{\mathrm{var}\left(x\right)}\\ =\sqrt{7.75}\\ \text{=2.78}\end{array}$

c) The binomial distribution is given as mentioned below.

$f\left(x\right)=C(n,x)p^x{q}^{n鈭�x}$

Where it's given the points mentioned:

$\begin{array}{c}n=48\\ \text{p=0.25}\\ \text{q=0.75}\\ \text{x=15}\end{array}$

Substitute the value.

$\begin{array}{c}f\left(15\right)=C(48,15){\left(0.25\right)}^{15}{\left(0.75\right)}^{33}\\ \text{=0.0767}\end{array}$

d) The normal distribution is given as mentioned below.

$f\left(x\right)=\frac{1}{\sqrt{2蟺npq}}{e}^{鈭�{(x鈭�渭)}^2/2npq}$

The probability is given as mentioned below.

$\begin{array}{c}f\left(15\right)=\frac{1}{\sqrt{\left(2蟺\right)\left(48\right)\left(0.25\right)\left(0.75\right)}}{e}^{鈭�0.5}\\ \text{=0.0806}\end{array}$

e) The Poisson distribution is given by

$f\left(x\right)=\frac{{\left(np\right)}^x{e}^{鈭�np}}{x!}$

Substitute in the above equation.

$f\left(x\right)=\frac{{\left(np\right)}^x{e}^{鈭�np}}{x!}\begin{array}{c}f\left(15\right)=\frac{{(48脳0.25)}^{15}}{15!}{e}^{鈭�12}\\ \text{=0.0724}\end{array}$

Hence, the solutions derived is mentioned below.

(a)

$\begin{array}{l}\text{p(x=2)=1/16,}\\ \text{p(x=4)=3/16,}\\ \text{p(x=6)=3/16,}\\ \text{p(x=8)=1/16}\end{array}$

(b)

$\begin{array}{l}渭=2.5,\\ 蟽=2.78\end{array}$

(c)$\text{Forbinomial}f\left(15\right)=0.0767$

(d)$\text{Fornormal}f\left(15\right)=0.0806$

(e)$\text{ForPoisson}f\left(15\right)=0.0724$