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The likelihood of an event happening is defined by probability. We may be required to predict the outcome of an event in a variety of real-life scenarios. The outcomes of an event may or may not be certain. In these instances, we state that the event has a chance of happening or not happening.

Denote

X= the number of diodes that failed on aboard.

X is binomial random variable with parameters

\(n = 200\)

and

\(p = 0.01,\)

both given in the exercise.

(a):

Proposition: For a binomial random variable $X$ with parameters $n, p$, and $q=1-p$, the following is true

\(\begin{array}{l}E(X) = np\\V(X) = np(1 - p) = npq\\{\sigma _X} = \sqrt {npq} \end{array}\)

The following is true

\(E(X) = np = 200 \cdot 0.01 = 2\)

also, the variance is

\(V(X) = npq = 200 \cdot 0.01 \cdot (1 - 0.01) = 1.98\)

from which we obtain the standard deviations as

\({\sigma _X} = \sqrt {V(X)} = \sqrt {1.98} = 1.407.\)

(b):

We need to find the approximate probability, therefore, look at the following proposition first.

Proposition: Assume that we have\(b(x;n,p)\)(pmf of binomial random variable), and that

\(np \to \mu > 0\)

when\(n \to \infty \)and\(p \to 0\), then\(b(x;n,p) \to p(x;\mu ),\)

where,\(p(x;\mu )\)is pmf of random variable with Poisson Distribution.

A random variable X with pmf

\(p(x;\mu ) = {e^{ - \mu }}\frac{{{\mu ^x}}}{{x!}}\)

for\(x = 0,1, \ldots \), is said to have Poisson Distribution with parameter\(\mu > 0\).

Since we have that

\(n = 200\)

and

\(p = 0.01,\)

the following is true

\(np = 200 \cdot 0.01 = 2.\)

Therefore,

\(\mu = 2\)

and X has approximately Poisson distribution with parameter\(\mu = 2\).

The following holds

\(\begin{array}{l}P(X \ge 4) = 1 - P(X < 4)\mathop = \limits^{(1)} 1 - P(X \le 3)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - F(3;2)\mathop = \limits^{(2)} 1 - 0.857\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.143\end{array}\)

(1): X can take only integer values;

(2): Appendix Table A. 2 contains the Poisson cdf\(F(x;\mu )\), you can always calculate the sum yourself.

(c):

Denote

Y= the number of board that that work properly among five.

Random variable X has Binomial Distribution with parameters

\(n = 5\)

because we have\(5\)boards, and

\({p_1} = P({\rm{ a board works properly }}).\)

For\({p_1}\), the following is true

\(\begin{array}{l}{p_1} = P({\rm{ a board works properly }})\\\mathop {\,\,\,\,\,\, = }\limits^{(1)} P(X = 0) = F(0;2)\;\;\;\\\,\,\,\,\,\mathop = \limits^{(2)} 0.135\end{array}\)

(1): all diodes have to work in order for board to work properly (zero diodes should fail);

(2): Appendix Table A. 2 contains the Poisson cdf\(F(x;\mu )\), you can always calculate yourself.

So, the random variable Y, as mentioned above, has Binomial distribution. Remember the following:

Theorem:

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

Considering the theorem, and the fact that parameters of the Binomial random variable Y are

\(n = 5\)

and

\(p = 0.135,\)

the following is true

\(\begin{array}{l}P(Y \ge 4)\mathop = \limits^{(1)} P(Y = 4) + P(Y = 5)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\begin{array}{*{20}{l}}5\\4\end{array}} \right){0.135^4} \cdot {(1 - 0.135)^{5 - 4}} + \left( {\begin{array}{*{20}{l}}5\\5\end{array}} \right){0.135^5} \cdot {(1 - 0.135)^{5 - 5}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.00144 + 0.00004\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.00148\end{array}\)