Chapter 4.8, Problem 4.93E

a.

To determine

To find:the expected number of sets required to complete the match for $P\left(A\right)=0.6$ .

Answer to Problem 4.93E

  $E\left(x\right)=4.0656$

Explanation of Solution

Given:

  $P\left(A\right)=0.6$

From previous exercise-4.92, it is found that $P\left(3\right)=0.28;P\left(4\right)=0.3744;P\left(5\right)=0.3456$ .

Calculation:

According to the question,

  $\begin{array}{l}E\left(x\right)={\displaystyle \sum _{x=3}^{5}p\left(x\right)}x\\ =0.28\left(3\right)+0.3744\left(4\right)+0.3456\left(5\right)\\ =0.84+1.4976+1.728\\ =4.0656\end{array}$

Hence, the expected number of sets required to complete the match is $E\left(x\right)=4.0656$ .

b.

To determine

To find: the expected number of sets required to complete the match under given conditions.

Answer to Problem 4.93E

  $E\left(x\right)=4.125$

Explanation of Solution

Given:

The given condition is:

When the players are of equal ability- that is $P\left(A\right)=0.5$

From previous exercise-4.92, it is found that $P\left(3\right)=0.28;P\left(4\right)=0.3744;P\left(5\right)=0.3456$ .

Calculation:

According to the question,

From previous exercise,

  $\begin{array}{l}p\left(3\right)=P{\left(A\right)}^3+{\left(1-P\left(A\right)\right)}^3\\ ={\left(0.5\right)}^3+{\left(0.5\right)}^3\\ =0.125+0.125\\ =0.25\end{array}$

  $\begin{array}{l}p\left(4\right)=C_2^{3}P{\left(A\right)}^3\left(1-P\left(A\right)\right)+C_2^3{\left(1-P\left(A\right)\right)}^{3}P\left(A\right)\\ =3{\left(0.5\right)}^4+3{\left(0.5\right)}^4\\ =0.1875+0.1875\\ =0.375\end{array}$

  $\begin{array}{l}p\left(5\right)=1-p\left(3\right)-p\left(4\right)\\ =1-0.25-0.375\\ =0.375\end{array}$

  $\begin{array}{l}E\left(x\right)={\displaystyle \sum _{x=3}^{5}p\left(x\right)}x\\ =0.25\left(3\right)+0.375\left(4\right)+0.375\left(5\right)\\ =0.75+1.5+1.875\\ =4.125\end{array}$

Hence, the expected number of sets required to complete the match is $E\left(x\right)=4.125$ .

c.

To determine

To find: the expected number of sets required to complete the match under given conditions.

Answer to Problem 4.93E

  $E\left(x\right)=3.3816$

Explanation of Solution

Given:

The given condition is:

When the players differ greatly in ability- that is $P\left(A\right)=0.9$

From previous exercise-4.92, it is found that $P\left(3\right)=0.28;P\left(4\right)=0.3744;P\left(5\right)=0.3456$ .

Calculation:

According to the question,

From previous exercise,

  $\begin{array}{l}p\left(3\right)=P{\left(A\right)}^3+{\left(1-P\left(A\right)\right)}^3\\ ={\left(0.9\right)}^3+{\left(0.1\right)}^3\\ =0.729+0.001\\ =0.73\end{array}$

  $\begin{array}{l}p\left(4\right)=C_2^{3}P{\left(A\right)}^3\left(1-P\left(A\right)\right)+C_2^3{\left(1-P\left(A\right)\right)}^{3}P\left(A\right)\\ =3{\left(0.9\right)}^3\left(0.1\right)+3{\left(0.1\right)}^3\left(0.9\right)\\ =0.2187+0.0027\\ =0.2214\end{array}$

  $\begin{array}{l}p\left(5\right)=1-p\left(3\right)-p\left(4\right)\\ =1-0.73-0.2214\\ =0.0486\end{array}$

  $\begin{array}{l}E\left(x\right)={\displaystyle \sum _{x=3}^{5}p\left(x\right)}x\\ =0.73\left(3\right)+0.2214\left(4\right)+0.0486\left(5\right)\\ =2.19+0.8856+0.243\\ =3.3816\end{array}$

Hence, the expected number of sets required to complete the match is $E\left(x\right)=3.3816$ .

d.

To determine

To find: the relation between $P\left(A\right)\text{ and }E\left(x\right)$ under the given condition.

Answer to Problem 4.93E

  $E\left(x\right)=4.0656$

Explanation of Solution

Given:

The given condition is:

The excepted number of sets required to complete the match. $P\left(A\right)=0.6$

Calculation:

According to the part (a) and part (b), it is observed that,

for $P\left(A\right)\ge 0.5$ ,

  $P\left(A\right)$ increases,

  $E\left(x\right)$ decreases.

Hence, the expected number of sets required to complete the match is $E\left(x\right)=4.0656$ .