91影视

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

a)

We are given \({\rm{10}}\) random variables (light bulbs). Each has an exponential distribution with parameter\({\rm{\lambda }}\).

Random variable \({\rm{X}}\) with pdf

is said to have exponential distribution with parameter\({\rm{\lambda }}\).

Let, we need to find the probability that all ten light bulbs fail before given time\({\rm{t}}\). This can be represented as intersection of \({\rm{10}}\) events

\({\rm{A = }}\bigcap\limits_{{\rm{i = 1}}}^{{\rm{10}}} {\left\{ {{{\rm{X}}_{\rm{i}}}{\rm{ < t}}} \right\}} \)

where \({{\rm{X}}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,10}}\) are the \({\rm{10}}\) random variables with exponential distribution.

Therefore, we have

\(\begin{array}{c}{\rm{P(A)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ < t}}} \right){\rm{ \times P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ < t}}} \right){\rm{ \times \ldots \times P}}\left( {{{\rm{X}}_{{\rm{10}}}}{\rm{ < t}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right)^{{\rm{10}}}}{\rm{,}}\end{array}\)

(1): multiplication property given below,

(2): cdf of exponentially distributed random variable is

Multiplication Property: Two events \({\rm{A}}\) and \({\rm{B}}\) are independent if and only if

\({\rm{P(A{C}B) = P(A) \times P(B)}}\)

b)

We need exactly \({\rm{k}}\) of \({\rm{10}}\) bulbs to fail before time\({\rm{t}}\), therefore, we want to define a random variable with Binomial Distribution.

The Binomial Random Variable \({\rm{X}}\) is defined as

\({\rm{X = }}\)the number of S's a month \({\rm{n}}\) trials,

where conditions one to four on page \({\rm{117}}\) are satisfied (binomial experiment).

Theorem:

\({\rm{b(x;n,p) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}&{{\rm{,x = 0,1,2, \ldots ,n}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

In our case, a success would be event

\({\rm{S = \{ bulb fails before time t\} ,}}\)

and the probability of such event is

Now we have random variable \({\rm{X}}\) which has Binomial Distribution with parameters \({\rm{n = 10}}\) and\({\rm{p}}\). Using the given theorem, it is easy to obtain the requested probability. Let, the following is true

\({\rm{P(X = k) = b}}\left( {{\rm{k;10,1 - }}{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right){\rm{ = }}\left( {\begin{array}{*{20}{c}}{{\rm{10}}}\\{\rm{k}}\end{array}} \right){\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right)^{\rm{k}}}{\left( {{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right)^{{\rm{n - k}}}}\)

c)

Similarly, as in\({\rm{(b)}}\), we want to create a situation where it is easy to use already calculated probabilities such as

\(\begin{array}{l}{\rm{P(S) = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ < t}}} \right){\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - \lambda t}}}}\\{\rm{P(F) = }}{{\rm{e}}^{{\rm{ - \lambda t}}}}\end{array}\)

Where the distribution parameter for the remaining light bulb's lifetime is\({\rm{\theta }}\).

Therefore, the probability that exactly five of the ten fails can be represented as union of two following events

\({{\rm{A}}_{\rm{1}}}{\rm{ = \{ 5}}\)with same parameter fail and other \({\rm{5}}\) do not fail \({\rm{\} }}\)

\({{\rm{A}}_{\rm{2}}}{\rm{ = }}\left\{ {\rm{4}} \right.\)with same parameter fail and \({{\rm{\theta }}^{\rm{鈯剗}}{\rm{s}}\) fail and other \({\rm{5}}\) do not fail \({\rm{\} }}\)

Define event

\({{\rm{E}}_{\rm{5}}}{\rm{ = \{ exactly 5 fail before time t\} }}\)

Hence, we have

\(\begin{aligned}{\rm{P}}\left( {{{\rm{E}}_{\rm{5}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{`E }}{{\rm{A}}_{\rm{2}}}} \right)\\{\rm{ = P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \left( {\begin{array}{*{20}{l}}{\rm{9}}\\{\rm{5}}\end{array}} \right){\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right)^{\rm{5}}}{\left( {{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right)^{\rm{4}}}{{\rm{e}}^{{\rm{ - \theta t}}}}{\rm{ + }}\left( {\begin{array}{*{20}{l}}{\rm{9}}\\{\rm{4}}\end{array}} \right){\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right)^{\rm{4}}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right){\left( {{{\rm{e}}^{{\rm{ - \lambda t}}}}} \right)^{\rm{5}}}\end{aligned}\)

(1): the same logic as in\({\rm{(b)}}\). We also know probabilities of each success/fail.