91Ó°ÊÓ

Probability denotes the possibility of something happening. It's a field of mathematics that studies the probability of a random event occurring.

a)

Denote random variable X as

X=number of classified as "seconds".

There are six goblets, so n=6, and 10% have the flaws and are classified as "seconds", which makes \(p = 0.1\). Therefore, \(X\sim {\mathop{\rm Bin}\nolimits} (6,0.1)\).

Theorem:

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

The following is true

\(\begin{aligned}b(1;6,0.1) &= P(X = 1)\\ &= \left( {\begin{array}{*{20}{l}}6\\1\end{array}} \right){0.1^1}{(1 - 0.1)^{6 - 1}}\\ &= 0.3543\end{aligned}\)

Therefore, the answer is \(0.3543\).

b)

Cumulative Density Function cdf of binomial random variable X with parameters n and p is

\(\begin{aligned}B(x;n,p) &= P(X \le x)\\ &= \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\end{aligned}\)

The probability is the following

\(\begin{aligned}P(X \ge 2)\mathop = \limits^{(1)} 1 - P\\(X < 2)\mathop = \limits^{(2)} 1 - P\\(X \le 1) = 1 - B\\(1;6,0.1) &= 1 - b\\(1;6,0.1) - b\\(0;6,0.1) &= 1 - 0.5313 - 0.3543\\& = 0.1143\end{aligned}\)

Therefore, the complement of event \(\{ X \ge 2\} \) is event \(\{ X < 2\} \);

As a result, the events are the same because X can take only non negative values.

c)

The event given can be represented as union of two disjoint event - either first four goblets does not have any flaws, or one of the first four does have a flaw is classified as second and fifth does not have a flaw.

The probability that first four goblets are without flaw is

\(\begin{array}{c}P(X = 0)\\\mathop = \limits^{(1)} \left( {\begin{array}{*{20}{l}}4\\0\end{array}} \right){0.1^0}{(1 - 0.1)^{4 - 0}}\\ = 0.6561\end{array}\)

(1) : out of four goblets, we want all to have a flaw.

The probability of the second given event is

\(\begin{array}{l}{410.1^1}{(1 - 0.1)^{4 - 1}} \cdot \\0.9 = 0.26244\end{array}\)

where the part

\(\left( {\begin{array}{*{20}{l}}4\\1\end{array}} \right){0.1^1}{(1 - 0.1)^{4 - 1}}\)

is the probability of selecting one goblet with a flaw out of four goblets and $0.9$ is the probability for fifth goblet to be good.

Therefore, probability of given event in the exercise is union of two disjoint events, its probability will be the sum of those two events, therefore, the probability of the event is \(0.6561 + 0.26244 = 0.91854\).