91影视

The amount of 'Success' in a series of n experiments, where each time a yes-no question is given, the Boolean-valued outcome is represented either with success/yes/true/one (probability\({\rm{p}}\)) or failure/no/false/zero (probability\({\rm{q = 1 - p}}\)) in a binomial probability distribution.

(a)

If an experiment satisfies conditions 鈥�

1. The experiment consists of a sequence of \({\rm{n}}\) smaller experiments called trials, where \({\rm{n}}\) is fixed in advance of the experiment.

2. Each trial can result in one of the same two possible outcomes (dichotomous trials), which generically is denote by success \({\rm{(S)}}\) and failure \({\rm{(F)}}\). The assignment of the \({\rm{S}}\) and \({\rm{F}}\) labels to the two sides of the dichotomy is arbitrary.

3. The trials are independent, so that the outcome on any particular trial does not influence the outcome on any other trial.

4. The probability of success \({\rm{P(S)}}\) is constant from trial to trial; this probability is denoted by \({\rm{p}}\).

It is called a binomial experiment.

The probability of success is not constant for each trial; therefore, the experiment cannot be binomial, and the random variable does not have binomial distribution.

Therefore, there is no binomial distribution.

(b)

Since it is possible for a false-positive and false-negative results, the event that exactly three of the ten results are positive will be union of four disjoint events. The four combinations are 鈥�

\({{\rm{A}}_{\rm{3}}}{\rm{ = 3}}\)with disease and positive, \({{\rm{B}}_{\rm{0}}}{\rm{ = 0}}\) without the disease and positive;

\({{\rm{A}}_2}{\rm{ = 2}}\)with disease and positive, \({{\rm{B}}_{\rm{1}}}{\rm{ = 1}}\) without the disease and positive;

\({{\rm{A}}_1}{\rm{ = 1}}\)with disease and positive, \({{\rm{B}}_{\rm{2}}}{\rm{ = 2}}\) without the disease and positive;

\({{\rm{A}}_{\rm{0}}}{\rm{ = 0}}\)with disease and positive, \({{\rm{B}}_{\rm{3}}}{\rm{ = 3}}\) without the disease and positive;

If it is denoted with 鈥�

\({\rm{C = \{ exactly 3 out of 10 tests are positive\} }}\)

Then it is obtained 鈥�

\(\begin{aligned}P(C) &= P\left( {\left\{ {{{\rm{A}}_{\rm{3}}} \cap {{\rm{B}}_{\rm{0}}}} \right\} \cup \left\{ {{{\rm{A}}_{\rm{2}}} \cap {{\rm{B}}_{\rm{1}}}} \right\} \cup \left\{ {{{\rm{A}}_{\rm{1}}} \cap {{\rm{B}}_{\rm{2}}}} \right\} \cup \left\{ {{{\rm{A}}_{\rm{0}}} \cap {{\rm{B}}_{\rm{3}}}} \right\}} \right)\\ &= P \left( {\left\{ {{{\rm{A}}_{\rm{3}}} \cap {{\rm{B}}_{\rm{0}}}} \right\}} \right){\rm{ + P}}\left( {\left\{ {{{\rm{A}}_{\rm{2}}} \cap {{\rm{B}}_{\rm{1}}}} \right\}} \right){\rm{ + P}}\left( {\left\{ {{{\rm{A}}_{\rm{1}}} \cap {{\rm{B}}_{\rm{2}}}} \right\}} \right){\rm{ + P}}\left( {\left\{ {{{\rm{A}}_{\rm{0}}} \cap {{\rm{B}}_{\rm{3}}}} \right\}} \right)\\ &= P\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ \times P}}\left( {{{\rm{B}}_{\rm{0}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ \times P}}\left( {{{\rm{B}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ \times P}}\left( {{{\rm{B}}_{\rm{2}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{ \times P}}\left( {{{\rm{B}}_{\rm{3}}}} \right)\end{aligned}\)

There are total of \({\rm{5}}\) whom have the disease and \({\rm{5}}\) of whom do not. And the probabilities of false-positive reading is \({{\rm{p}}_{\rm{1}}}{\rm{ = 0}}{\rm{.2}}\) and the probability of false-negative reading is \({{\rm{p}}_{\rm{2}}}{\rm{ = 0}}{\rm{.1}}\).

For \({\rm{i = 1,2,3}}\), it is obtained 鈥�

\({\rm{P(}}{{\rm{A}}_{\rm{i}}}{\rm{) = }}\left( {\begin{array}{*{20}{c}}{\rm{5}}\\{\rm{i}}\end{array}} \right){\rm{p}}_{\rm{1}}^{\rm{i}}{{\rm{(1 - }}{{\rm{p}}_{\rm{1}}}{\rm{)}}^{{\rm{5 - i}}}}\)

Also, for \({\rm{i = 1,2,3}}\), it is obtained 鈥�

\({\rm{P(}}{{\rm{B}}_{\rm{i}}}{\rm{) = }}\left( {\begin{array}{*{20}{c}}{\rm{5}}\\{\rm{i}}\end{array}} \right){\rm{p}}_{\rm{2}}^{{\rm{5 - i}}}{{\rm{(1 - }}{{\rm{p}}_{\rm{2}}}{\rm{)}}^{\rm{i}}}\)

For the given \({{\rm{p}}_{\rm{1}}}\) and \({{\rm{p}}_{\rm{2}}}\), the first term is given with 鈥�

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{0}}}} \right){\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{3}}\end{array}} \right){\rm{p}}_{\rm{1}}^{\rm{3}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)^{{\rm{5 - 3}}}} \cdot \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{0}}\end{array}} \right){\rm{p}}_{\rm{2}}^{{\rm{5 - 0}}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)^{\rm{0}}}\\{\rm{ = 0}}{\rm{.0512}} \cdot {\rm{0}}{\rm{.00001 = 0}}{\rm{.000000512}}\end{array}\)

The second term with 鈥�

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{1}}}} \right){\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right){\rm{p}}_{\rm{1}}^{\rm{2}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)^{{\rm{5 - 2}}}} \cdot \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right){\rm{p}}_{\rm{2}}^{{\rm{5 - 1}}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)^{\rm{1}}}\\{\rm{ = 0}}{\rm{.2048}} \cdot {\rm{0}}{\rm{.00045 = 0}}{\rm{.00009216}}\end{array}\)

The third term with 鈥�

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{2}}}} \right){\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{1}}\end{array}} \right){\rm{p}}_{\rm{1}}^{\rm{1}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)^{{\rm{5 - 1}}}} \cdot \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{2}}\end{array}} \right){\rm{p}}_{\rm{2}}^{{\rm{5 - 2}}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)^{\rm{2}}}\\{\rm{ = 0}}{\rm{.4096}} \cdot {\rm{0}}{\rm{.0081 = 0}}{\rm{.00332}}\end{array}\)

The fourth term with 鈥�

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{3}}}} \right){\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{0}}\end{array}} \right){\rm{p}}_{\rm{1}}^{\rm{0}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)^{{\rm{5 - 0}}}} \cdot \left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{3}}\end{array}} \right){\rm{p}}_{\rm{2}}^{{\rm{5 - 3}}}{\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)^{\rm{3}}}\\{\rm{ = 0}}{\rm{.32768}} \cdot {\rm{0}}{\rm{.0729 = 0}}{\rm{.02389}}\end{array}\)

Then finally 鈥�

\(\begin{aligned}P(C) &= P \left( {{{\rm{A}}_{\rm{3}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{0}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{1}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{2}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{0}}}} \right) \cdot {\rm{P}}\left( {{{\rm{B}}_{\rm{3}}}} \right)\\&= 0 {\rm{.000000512 + 0}}{\rm{.00009216 + 0}}{\rm{.00332 + 0}}{\rm{.02389}}\\&= 0 {\rm{.027302672}}\end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.027302672}}\).