91Ó°ÊÓ

The probability of scoring 20 on a throw of the dart.

Given: Three circles, each with a radius of 1,2,3, are located in the centre of a board.

The circle of radius 1 scoring points =30.

The circle of radius 2 scoring points =20.

The circle of radius 3 scoring points =10.

Calculation:

$\mathrm{P}\left(1\right)=\frac{1}{3},\mathrm{P}\left(2\right)=\frac{1}{3},\mathrm{P}\left(3\right)=\frac{1}{3}$

Let A}be the score 20 on a throw of a dart $\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{3}$

The probability of scoring at least 20 on a throw of the dart.

Given : $\mathrm{P}\left(1\right)=\frac{1}{3},\mathrm{P}\left(2\right)=\frac{1}{3},\mathrm{P}\left(3\right)=\frac{1}{3}$

Calculation:

Let B be the score at least 20 on a throw of a dart

$$\begin{gathered} P\left(\mathrm{B}\right)=\mathrm{P}\left(2\right)+\mathrm{P}\left(3\right) \\ =\frac{1}{3}+\frac{1}{3} \\ P\left(B\right)=\frac{2}{3} \end{gathered}$$

The probability of scoring 0 on a throw of a dart.

Given: $\mathrm{P}\left(1\right)=\frac{1}{3},\mathrm{P}\left(2\right)=\frac{1}{3},\mathrm{P}\left(3\right)=\frac{1}{3}$

Calculation:

Let C be the score 0 on a throw of a dart

$\mathrm{P}\left(\mathrm{C}\right)=\frac{0}{3}=0$

The expected value of a score on a throw of a dart.

Given: $\mathrm{P}\left(1\right)=\frac{1}{3},\mathrm{P}\left(2\right)=\frac{1}{3},\mathrm{P}\left(3\right)=\frac{1}{3}$

Calculation:

Let D be the expected value of a score on a throw of a dart

$E\left(D\right)=\frac{\text{sum of total points}}{\text{number of throw}}=\frac{10+20+30}{3}=\frac{60}{3}=20$

Hence $E\left(D\right)=20$

The first two throw score at least 10 .

Given: $\mathrm{P}\left(1\right)=\frac{1}{3},\mathrm{P}\left(2\right)=\frac{1}{3},\mathrm{P}\left(3\right)=\frac{1}{3}$

Calculation: Let E be first two throw score at least 10 .
$\begin{array}{r}\mathrm{P}\left(\mathrm{E}\right)=\left(\mathrm{P}\right(1\left)\mathrm{P}\right(2\left)\right)+\left(\mathrm{P}\right(1\left)\mathrm{P}\right(3\left)\right)+\left(\mathrm{P}\right(2\left)\mathrm{P}\right(1\left)\right)+\left(\mathrm{P}\right(2\left)\mathrm{P}\right(3\left)\right)+\left(\mathrm{P}\right(3\left)\mathrm{P}\right(1\left)\right)+\left(\mathrm{P}\right(3\left)\mathrm{P}\right(2\left)\right)\\ +\left(\mathrm{P}\right(1\left)\mathrm{P}\right(0\left)\right)+\left(\mathrm{P}\right(3\left)\mathrm{P}\right(0\left)\right)+\left(\mathrm{P}\right(2\left)\mathrm{P}\right(0)+(\mathrm{P}\left(0\right)\mathrm{P}\left(3\right))+(\mathrm{P}\left(0\right)\mathrm{P}\left(2\right))+(\mathrm{P}\left(0\right)\mathrm{P}\left(1\right))\\ =\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}\right.\\ ×0)+\left(\frac{1}{3}×0\right)+\\ \left(\frac{1}{3}+0\right)+\left(\frac{1}{3}×0\right)+\left(\frac{1}{3}×0\right)+\left(\frac{1}{3}×0\right)\\ =\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+0+0+0+0+0+0=\frac{6}{9}=\frac{2}{3}\end{array}$

Hence $P\left(E\right)=\frac{2}{3}$

To calculate total score after two throw is 30.

Given :$\mathrm{P}\left(1\right)=\frac{1}{3},\mathrm{P}\left(2\right)=\frac{1}{3},\mathrm{P}\left(3\right)=\frac{1}{3}$

Calculation: Let F be the score after two throw 30

$\begin{array}{r}\mathrm{P}\left(\mathrm{F}\right)=\left(\mathrm{P}\right(1\left)\mathrm{P}\right(2\left)\right)+\left(\mathrm{P}\right(3)+\mathrm{P}(0\left)\right)+\left(\mathrm{P}\right(2\left)\mathrm{P}\right(1\left)\right)+\left(\mathrm{P}\right(0)+\mathrm{P}(3\left)\right)\\ =\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}×0\right)+\left(\frac{1}{3}×\frac{1}{3}\right)+\left(\frac{1}{3}×0\right)\\ =\left(\frac{1}{9}\right)+0+\left(\frac{1}{9}\right)+0=\frac{2}{9}\end{array}$

Hence$P\left(F\right)=\frac{2}{9}$