Q.4.12 - A First Course in Probability

Question

A local soccer team has $5$ more games left to play. If it wins its game this weekend, then it will play its final $4$ games in the upper bracket of its league, and if it loses, then it will play its final games in the lower bracket. If it plays in the upper bracket, then it will independently win each of its games in this bracket with probability .$4$, and if it plays in the lower bracket, then it will independently win each of its games with probability .$7$. If the probability that the team wins its game this weekend is .$5$, what is the probability that it wins at least $3$ of its final $4$ games? 

Step-by-Step Solution

Answer

The required probability is $0.6965$ 

Step 1:Given Information

Given that the local soccer team has $5$ more games left to play. If it wins its game this weekend, then it will play its final $4$ games in the upper bracket of its league, and if it loses, then it will play its final games in the lower bracket. If it plays in the upper bracket, then it will independently win each of its games in this bracket with probability $.4$, and if it plays in the lower bracket, then it will independently win each of its games with probability .$7$. If the probability that the team wins its game this weekend is .$5$.

Step 2:Explanation

Describe $X$ as the random variable which calculates the number of matches that the local soccer group succeeds out of the last four matches. We want to compute $P(X\ge 3)$. If the first match has been succeeded we have that $X~Binom(4,0.4)$. If the first match has been failed, we are given that $X~Binom(4,0.7)$. Utilizing LOTP, we have that

$P(X\ge 3)=P(X\ge 3\mid \text{ first game won })P\left(\text{ first game won }\right)$

$+P(X\ge 3\mid \text{ first game lost })P\text{ (first game lost) }$

$=\frac{1}{2}\left(\left(\begin{array}{l}4\\ 3\end{array}\right){0.4}^{3}0.6+\left(\begin{array}{l}4\\ 4\end{array}\right){0.4}^4\right)+\frac{1}{2}\left(\left(\begin{array}{l}4\\ 3\end{array}\right){0.7}^{3}0.3+\left(\begin{array}{l}4\\ 4\end{array}\right){0.7}^4\right)$

$=0.6965$

Step 3:Final Answer

$0.6965$ is the probability that it wins at least $3$ of its final $4$ games.