Probability Practice Test 1

Concept:When two events compete in alternating turns, use a geometric series to find each player's winning probability.Explanation:Let $P(6)=\frac{5}{36}$ be the probability A throws a total of 6 with a pair of dice.Let $P(7)=\frac{6}{36}=\frac{1}{6}$ be the probability B throws a total of 7.The probability that neither wins in a round $=\ 1-\frac{5}{36}-\frac{6}{36}=\frac{25}{36}$.A wins on his first roll, or both fail and the situation repeats.So, $P(A)=\frac{5}{36}+\frac{25}{36}P(A)$.$\Rightarrow P(A)\left(1-\frac{25}{36}\right)=\frac{5}{36}$$\Rightarrow P(A)=\frac{5}{36}\times\frac{36}{11}=\frac{5}{11}$.Wait, checking alternately: A wins immediately with $\frac{5}{36}$; if A fails and B fails, the game restarts.$P(A)=\frac{5}{36}+\left(\frac{31}{36}\right)\left(\frac{30}{36}\right)P(A)$$P(A)=\frac{5}{36}\times\frac{1296}{366}=\frac{30}{61}$.Then $P(B)=1-\frac{30}{61}=\frac{31}{61}$.Answer:$P(A)=\frac{30}{61}$ and $P(B)=\frac{31}{61}$.