5 In this question, give probabilities correct to 4 decimal places.
A contestant in a game-show starts with one, two or three 'lives', and then performs a series of tasks. After each task, the number of lives either decreases by one, or remains the same, or increases by one. The game ends when the number of lives becomes either four or zero. If the number of lives is four, the contestant wins a prize; if the number of lives is zero, the contestant loses and leaves with nothing.
At the start, the number of lives is decided at random, so that the contestant is equally likely to start with one, two or three lives. The tasks do not involve any skill, and after every task:
- the probability that the number of lives decreases by one is 0.5 ,
- the probability that the number of lives remains the same is 0.05 ,
- the probability that the number of lives increases by one is 0.45 .
This is modelled as a Markov chain with five states corresponding to the possible numbers of lives. The states corresponding to zero lives and four lives are absorbing states.
- Write down the transition matrix \(\mathbf { P }\).
- Show that, after 8 tasks, the probability that the contestant has three lives is 0.0207 , correct to 4 decimal places.
- Find the probability that, after 10 tasks, the game has not yet ended.
- Find the probability that the game ends after exactly 10 tasks.
- Find the smallest value of \(N\) for which the probability that the game has not yet ended after \(N\) tasks is less than 0.01 .
- Find the limit of \(\mathbf { P } ^ { n }\) as \(n\) tends to infinity.
- Find the probability that the contestant wins a prize.
The beginning of the game is now changed, so that the probabilities of starting with one, two or three lives can be adjusted.
- State the maximum possible probability that the contestant wins a prize, and how this can be achieved.
- Given that the probability of starting with one life is 0.1 , and the probability of winning a prize is 0.6 , find the probabilities of starting with two lives and starting with three lives.
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