5. Rachel and Steven play games of chess. If either wins two consecutive games $s /$ he is declared the champion.
The probability that Rachel will win any given game is 0.6 .
The probability that Steven will win any given game is 0.3 .
There is a 0.1 probability that any given game is drawn.
The probability that neither is the champion after at most three games is $P$. Find the value of $1000 P$.
Solution
343
We use $R$ to denote the event 'Rachel wins a game'; $S$ to denote 'Steven wins a game'; and $D$ to denote 'a game is drawn'. We use $\bar{R}$ to denote the event 'Rachel does not win a game', with $\bar{S}$ defined similarly.
- Event $R, D$ has probability $0.6 \times 0.1=0.06$.
- Event $R, S, \bar{S}$ has probability $0.6 \times 0.3 \times 0.7=0.126$.
- Event $D, R, \bar{R}$ has probability $0.1 \times 0.6 \times 0.4=0.024$.
- Event $D, S, \bar{S}$ has probability $0.1 \times 0.3 \times 0.7=0.021$.
- Event $S, D$ has probability $0.3 \times 0.1=0.03$.
- Event $S, R, \bar{R}$ has probability $0.3 \times 0.6 \times 0.4=0.072$.
- Event $D, D$ has probability $0.1 \times 0.1=0.01$.

Therefore $P=0.06+0.126+0.024+0.021+0.03+0.072+0.01=0.343$; hence $1000 P=343$.