91影视

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

Define with event 鈥�

\({\rm{S = }}\){start-up is successful}

It is given that 鈥�

\({\rm{p = P(S)}}\)

Assume that a power-generating unit required \({\rm{2}}\) consecutive successful start-ups, where \({\rm{p = 0}}{\rm{.9}}\).

There can be at least two start-ups that must be made prior acceptance (first two successful), therefore the smallest value that random variable\({\rm{X}}\)can take is\({\rm{2}}\), and can take any integer bigger than\({\rm{2}}\).

Obviously, for\({\rm{x = 2}}\)it is given 鈥�

\(\begin{aligned}P(X = 2) &= P(SS) \\ &= pp = {p^2} \\ &= {(0.9)^2} = 0.81 \\ \end{aligned} \)

For\({\rm{x = 3}}\), since the last two start-ups have to be consecutive in order for unit to be accept the first start-up has to fail, therefore it is given 鈥�

\(\begin{aligned}P(X = 3) &= P(FSS) \\ & = (1 - p){p^2} \\ \end{aligned} \)

Generally, for\({\rm{x}} \ge {\rm{4}}\), the last three events have to be ordered as 鈥�

fail start-up, successful start-up, successful start-up,

Because when there are two consecutive successful start-ups the unit is accepted and the process is finished, so there cannot be\({\rm{2}}\)successful start-ups before the last\({\rm{2}}\).

In case when\({\rm{x = 4}}\), since we know the last three events,\({\rm{x - 3 = 4 - 3 = 1}}\)event could be any other event. It can be written as 鈥�

\(\begin{gathered}P(X = 4) &= P(FFSS \cup E SFSS) \\ &= (1 - {p^2}) + (1 - p){p^3} \\\end{gathered} \)

For \({\rm{x}} \ge {\rm{5}}\), there are \({\rm{x - 3 = 5 - 3}} \ge {\rm{2}}\) events before the last three. Find the probability that the first \({\rm{2}}\) or \({\rm{3}}\) or ... or \({\rm{x - 3}}\) events do not have \({\rm{2}}\) consecutive successful start-ups, and that is intersection of events 鈥�

\({\rm{\{ X = 2\} ',\{ X = 3\} ',}}...{\rm{,\{ X = x - 3\} '}}\)

Where, for example, event \({\rm{\{ X = x - 3\} '}}\) is the complement of event that in the first \({\rm{x - 3}}\) events there are \({\rm{2}}\) consecutive successful start-ups.

The following is true 鈥�

\(\begin{aligned}P\left( {\{ X = 2\} ' \cap \{ X = 3\} ' \cap \ldots \cap \{ X = x - 3\} '} \right) \\ & = 1 - P\left[ {\left\{ {\{ X = 2\} ' \cap \{ X = 3\} ' \cap \ldots \cap \{ X = x - 3\} '} \right\}'} \right] \\ &= 1 - P[\{ X = 2\} \cup \{ X = 3\} \cup \ldots \cup \{ X = x - 3\} ] \\ &= 1 - [P(X = 2) + P(X = 3) + \ldots + P(X = x - 3)] \\ \end{aligned} \)

Finally, because of the independence of outcomes, we can express \({\rm{P(X = x)}}\) for \({\rm{x}} \ge {\rm{5}}\) "recursively" as 鈥�

\({\rm{P(X = x) = (1 - (P(X = 2) + P(X = 3) + \ldots + P(X = x - 3)))}} \cdot {\rm{(1 - p)}}{{\rm{p}}^{\rm{2}}}\)

Where \({\rm{(1 - p)}}{{\rm{p}}^{\rm{2}}}\) stands for the probability of event \(FSS\) as mentioned before.

Using the given probability \({\rm{p = 0}}{\rm{.9}}\), compute the only probabilities 鈥�

\({\rm{P(X = x), x = 2,3,}}...{\rm{,8,}}\)

because we need to find \({\rm{P(X}} \le {\rm{8)}}\) which is the sum of such probabilities. Therefore, manually computing all \({\rm{7}}\) probabilities, we get the following \({\rm{pmf}}\) of \({\rm{X}}\) for the first \({\rm{7}}\) values 鈥�

The requested probability is 鈥�

\(\begin{aligned}P(X \leqslant 8) &= P(X = 2) + P(X = 3) + \ldots + P(X = 8) \\ &= 0.81 + 0.081 + 0.081 + 0.0154 + 0.0088 + 0.0023 + 0.001 \\ & = 0.9995 \\ \end{aligned} \)

Therefore, the value is obtained as \({\rm{0}}{\rm{.9995}}\).