Q.4.73 - A First Course in Probability

Question

 Suppose in Problem 4.72 that the two teams are evenly matched and each has probability 1 2 of winning each game. Find the expected number of games played.

Step-by-Step Solution

Answer

In the given information the answer is $EX=5.8125$

Step 1 : Given Information

Given that the two players are evenly matched, evenly matched means the two teams are played same number of matches with equal probability.

The probability is$\frac{1}{2}$.

By the definition $E\left(X\right)=\sum xp\left(x\right)$.

Let E games played=$E\left[GP\right]$

Expected value is $E\left(GP\right)=2\sum xp\left(x\right)$.

Step 2 : Calculation

$E\left[\mathrm{GP}\right]=2\left\{\begin{array}{l}4\times \left(\begin{array}{l}3\\ 3\end{array}\right){\left(\frac{1}{2}\right)}^4+5\left[\frac{1}{2}\times {}^{4}C_3{\left(\frac{1}{2}\right)}^3\left(\frac{1}{2}\right)\right]+6\left[\frac{1}{2}\times {}^{3}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^2\right]\\ +7\left[\frac{1}{2}\times {}^{6}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^3\right]\end{array}\right\}$

$E\left[\mathrm{GP}\right]=2\left\{\frac{1}{4}+5\left(\frac{1}{8}\right)+6\left(\frac{5}{32}\right)+7\times \left(\frac{5}{32}\right)\right\}$

            $$\begin{gathered} =\frac{2\times 93}{32} \\ =5.8125 \end{gathered}$$

Step 3:Final Answer

The final answer is  $EX$=$5.8125$