91Ó°ÊÓ

Probability is a metric for determining the possibility of an event occurring. Many things are impossible to forecast with \({\rm{100\% }}\)accuracy. Using it, we can only anticipate the probability of an event occurring, or how probable it is to occur. Probability can range from \({\rm{0}}\) to \({\rm{1}}\), with \({\rm{0}}\) indicating an improbable event and 1 indicating a certain event. Possibility of...

(a) It is specified \({\rm{Y}}\)as the distance from the left end of the bar at which it snaps. The beta distribution of \({\rm{Y/20}}\)is standard:

\(\begin{array}{l}{\rm{E(Y) = 10}}\\{\rm{V(Y) = }}\frac{{{\rm{100}}}}{{\rm{7}}}\end{array}\)

Let us take a look at following proposition:

Proposition : When \({\rm{h(X) = aX + b}}\), then \({\rm{h(X)}}\) anticipated value and standard deviation meet the following criteria.

\(\begin{aligned}E[h(X)] &= aE[X] + b \\{\sigma _{h(x)}} &= a{\sigma _x} \\\end{aligned} \)

We may write expected value and variance of using the above proposition\({\rm{Y/20}}\):

\(\begin{aligned}E\left( {\frac{Y}{{20}}} \right) &= \frac{1}{{20}} \times 10\\&= \frac{1}{2} \\ V\left( {\frac{Y}{{20}}}\right) &= \frac{1}{{{{20}^2}}} \times \frac{{100}}{7}\\&= \frac{1}{{28}} \\\end{aligned}\)

Since it is given that \({\rm{Y/20}}\) has a standard beta distribution then in terms of parametes \({\rm{\alpha }}\)and \({\rm{\beta }}\), expressions of it's expected value is :

\(\begin{aligned}E(X)&= \frac{\alpha }{{\alpha + \beta }} \\\frac{1}{2} &= \frac{\alpha }{{\alpha + \beta }} \\\alpha + \beta &= 2\alpha \\\beta &= \alpha \\\end{aligned} \)

Similarly variance \({\rm{V(X)}}\)can be written as:

\(\begin{aligned}V(X)&= \frac{{\alpha \beta }}{{{{(\alpha+\beta)}^2}(\alpha+\beta+ 1)}}\\\frac{5}{7}&= \frac{{\alpha \beta }}{{{{(\alpha+\beta)}^2}(\alpha+\beta+ 1)}}\\\end{aligned}\)

Now we make use of (eq.l) here :

\(\begin{aligned}\frac{1}{{28}} &= \frac{{\alpha \alpha }}{{{{(\alpha + \alpha )}^2}(\alpha + \alpha + 1)}} \\\frac{1}{{28}} &= \frac{{{\alpha ^2}}}{{\left( {4{\alpha ^2}} \right)(2\alpha + 1)}} \\ 4(2\alpha + 1) &= 28 \\\alpha &= 3 \\\beta &= 3 \\\end{aligned} \)

Therefore, The joint is \({\rm{\alpha = \beta = 3}}\).

(b) The standard beta distribution's pdf is as follows (let us denote \({\rm{Y/20}}\)as rv \({\rm{X}}\)):

\({\rm{f(x,\alpha ,\beta ) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\Gamma ({\rm{\alpha + \beta )}}}}{{\Gamma (\alpha {\rm{) \times }}\Gamma (\beta )}}{\rm{ \times }}{{\rm{x}}^{{\rm{\alpha - 1}}}}{\rm{ \times (1 - x}}{{\rm{)}}^{{\rm{\beta - 1}}}}}&{0 \le x \le 1}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

Substituting the parameters' values \({\rm{\alpha = 3}}\)and \({\rm{\beta = 3}}\), we get:

\({\rm{f(x,3,3) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{30 \times }}{{\rm{x}}^{\rm{2}}}{\rm{ \times (1 - x}}{)^2}}&{0 \le x \le 1}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

The \({\rm{P(8}} \le {\rm{Y}} \le 12)\)can be written as :

\(\begin{aligned}P(8 \leqslant Y \leqslant 12)&= P\left( {\frac{8}{{20}} \leqslant \frac{Y}{{20}} \leqslant \frac{{12}}{{20}}} \right) \\&= P(0.4 \leqslant X \leqslant 0.6) \\&= \int_{0.4}^{0.6} f (x) \times dx \\&= \int_{0.4}^{0.6} 3 0 \times {x^2} \times {(1 - x)^2} \times dx \\&= \int_{0.4}^{0.6} 3 0 \times {x^2} \times \left( {{x^2} - 2x + 1} \right) \times dx \\&= 30\int_{0.4}^{0.6} {\left( {{x^4} - 2{x^3} + {x^2}} \right)} \times dx \\\end{aligned}\)

\(\begin{aligned}&= 30\left[ {\frac{{{x^5}}}{5} - 2\frac{{{x^4}}}{4} + \frac{{{x^3}}}{3}} \right]_{0.4}^{0.6} \hfill \\&= 30\left[ {\frac{{{x^5}}}{5} - \frac{{{x^4}}}{2} + \frac{{{x^3}}}{3}} \right]_{0.4}^{0.6} \hfill \\&= 30\left[ {\left( {\frac{{{{(0.6)}^5}}}{5} - \frac{{{{(0.6)}^4}}}{2} + \frac{{{{(0.6)}^3}}}{3}} \right) - \left( {\frac{{{{(0.4)}^5}}}{5} - \frac{{{{(0.4)}^4}}}{2} + \frac{{{{(0.4)}^3}}}{3}} \right)} \right] \hfill \\&= 30[0.02275 - 0.01058] = 0.3651 \hfill \\\end{aligned}\)

\({\rm{P(}}8 \le Y \le 12) = 0.3651\)

Therefore, The probability is \(0.3651\)

(c) We expect it to snap at \({\rm{10}}\) inches since \({\rm{E(Y) = 10}}\). As a result, the likelihood that the bar snaps more than \({\rm{2}}\) inches from where we expect it to can be expressed as \({\rm{P(X}} < 8\)or \({\rm{Y}} > 12)\).

\(\begin{aligned}P(X < 8{\text{ }}or{\text{ }}Y > 12) & = 1 - P(8 \leqslant Y \leqslant 12) \\ &= 1 - 0.3651P(X < 8{\text{ }}or{\text{ }}Y > 12) \\&= 0.6349 \\\end{aligned}\)

Therefore, The probability is \(0.6349\)