In a tournament, a team plays 10 matches with probabilities of winning and losing each mat…

Question

In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $$\frac{1}{3}$$ and $$\frac{2}{3}$$ respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that t...

Answer: 8288

Accepted Answer

$ $$\& x+y=10  \  \& A=x-y  \  \& P(|A|(1) $A=0\Rightarrow x=5=y$$$

$P(A=0)={}\^{10}{C}\_{5}{(\frac{1}{3})}\^{5}{(\frac{2}{3})}\^{5}$

(2) $A=-2$

$\Rightarrow x=4$ and $y=6$

$P(A=-2)={}\^{10}{C}\_{4}\cdot {(\frac{1}{3})}\^{4}{(\frac{2}{3})}\^{6}$ and

$$$\begin{gathered} P(A=2)={}^{10}{C}_{6}{(\frac{1}{3})}^{6}{(\frac{2}{3})}^{4} \ \Rightarrow P(|A|\le 2){3}^{9}=3({}^{10}{C}_{5}\cdot {2}^{5}+{}^{10}{C}_{4}\cdot {2}^{6}+{}^{10}{C}_{6}\cdot {2}^{4}) \ =8288 \end{gathered}$$

Solutions · 1