Difference between revisions of "Probability RV Exercises"
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= Probability Exercises = | = Probability Exercises = | ||
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+ | Levels in [square brackets] and solutions in {curly brackets}. Worked solution clips can be found here [http://youtu.be/GsHP3gMwqI8?hd=1 Q1], [http://youtu.be/4zSRlomN6q4?hd=1 Q2], [http://youtu.be/GnJ0_DEymxI?hd=1 Q3], [http://youtu.be/lFkoxN4QO0M?hd=1 Q4] and [http://youtu.be/4MwFJog0kSo?hd=1 Q5]. | ||
<ol> | <ol> | ||
− | <li><p>In an experiment, if a mouse is administered dosage level <math>A</math> of a certain (harmless) hormone then there is a <math>0.2</math> probability that the mouse will show signs of aggression within one minute. For dosage levels <math>B</math> and <math>C</math>, the probabilities are <math>0.5</math> and <math>0.8</math>, respectively. Ten mice are given exactly the same dosage level of the hormone and, of these, exactly <math>6</math> shows signs of aggression within one minute of receiving the dose.</p> | + | <li><p><math>[L1,L2]</math> In an experiment, if a mouse is administered dosage level <math>A</math> of a certain (harmless) hormone then there is a <math>0.2</math> probability that the mouse will show signs of aggression within one minute. For dosage levels <math>B</math> and <math>C</math>, the probabilities are <math>0.5</math> and <math>0.8</math>, respectively. Ten mice are given exactly the same dosage level of the hormone and, of these, exactly <math>6</math> shows signs of aggression within one minute of receiving the dose.</p> |
<ol> | <ol> | ||
− | <li><p>Calculate the probability of this happening for each of the three dosage levels, <math>A,B</math> and <math>C</math>. (This is essentially a Binomial random variable problem, so you can check your answers using EXCEL.)</p></li> | + | <li><p>Calculate the probability of this happening for each of the three dosage levels, <math>A,B</math> and <math>C</math>. (This is essentially a Binomial random variable problem, so you can check your answers using EXCEL.) {0.0055; 0.2051; 0.0881}</p></li> |
− | <li><p>Assuming that each of the three dosage levels was equally likely to have been administered in the first place (with a probability of <math>1/3</math>), use Bayes’ Theorem to evaluate the likelihood of each of the dosage levels ''given'' that <math>6</math> out of the <math>10</math> mice were observed to react in this way.</p></li></ol> | + | <li><p>Assuming that each of the three dosage levels was equally likely to have been administered in the first place (with a probability of <math>1/3</math>), use Bayes’ Theorem to evaluate the likelihood of each of the dosage levels ''given'' that <math>6</math> out of the <math>10</math> mice were observed to react in this way. Most likely that dosis B was administered.</p></li></ol> |
</li> | </li> | ||
− | <li><p>Let <math>X</math> be the random variable indicating the number of incoming planes every <math>k</math> minutes at a large international airport, with probability mass function given by <math>p(x)=\Pr (X=x)=\frac{(0.9k)^{x}}{x!}\exp (-0.9k),\quad x=0,1,2,3,4,..</math>. Find the probabilities that there will be</p> | + | <li><p><math>[L1,L2]</math> Let <math>X</math> be the random variable indicating the number of incoming planes every <math>k</math> minutes at a large international airport, with probability mass function given by <math>p(x)=\Pr (X=x)=\frac{(0.9k)^{x}}{x!}\exp (-0.9k),\quad x=0,1,2,3,4,..</math>. Find the probabilities that there will be</p> |
<ol> | <ol> | ||
− | <li><p>exactly <math>9</math> incoming planes during a period of <math>5</math> minutes (i.e., find <math>\Pr (X=9)</math> when <math>k=5)</math>;</p></li> | + | <li><p>exactly <math>9</math> incoming planes during a period of <math>5</math> minutes (i.e., find <math>\Pr (X=9)</math> when <math>k=5)</math>; { 0.02316}</p></li> |
− | <li><p>fewer than <math>5</math> incoming planes during a period of <math>4</math> minutes (i.e., find <math>\Pr (X<5)</math> when <math>k=4)</math>;</p></li> | + | <li><p>fewer than <math>5</math> incoming planes during a period of <math>4</math> minutes (i.e., find <math>\Pr (X<5)</math> when <math>k=4)</math>; { 0.7065}</p></li> |
− | <li><p>at least <math>4</math> incoming planes during an <math>2</math> minute period (i.e., find <math>\Pr (X\geq 4)</math> when <math>k=2)</math>.</p></li></ol> | + | <li><p>at least <math>4</math> incoming planes during an <math>2</math> minute period (i.e., find <math>\Pr (X\geq 4)</math> when <math>k=2)</math>. { 0.1087}</p></li></ol> |
<p>Check all your answers using EXCEL.</p></li> | <p>Check all your answers using EXCEL.</p></li> | ||
− | <li><p>The random variable <math>Y</math> is said to be ''Geometric'' if it has probability mass function given by <math>p(y)=\Pr (Y=y)=(1-\theta )\theta ^{y-1},\quad y=1,2,3,...;\quad 0<\theta | + | <li><p><math>[L2]</math> The random variable <math>Y</math> is said to be ''Geometric'' if it has probability mass function given by <math>p(y)=\Pr (Y=y)=(1-\theta )\theta ^{y-1},\quad y=1,2,3,...;\quad 0<\theta |
<1</math>; where <math>\theta </math> is an unknown ‘parameter’.</p> | <1</math>; where <math>\theta </math> is an unknown ‘parameter’.</p> | ||
<p>Show that the cumulative distribution function can be expressed as</p> | <p>Show that the cumulative distribution function can be expressed as</p> | ||
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+\theta ^{y-1}\right) .</math></p> | +\theta ^{y-1}\right) .</math></p> | ||
<p>The term in the second bracket on the right-hand side is the sum of a ''Geometric Progression''.)</p></li> | <p>The term in the second bracket on the right-hand side is the sum of a ''Geometric Progression''.)</p></li> | ||
− | <li><p>The weekly consumption of fuel for a certain machine is modelled by means of a continuous random variable, <math>X</math>, with probability density function</p> | + | <li><p><math>[L2]</math> The weekly consumption of fuel for a certain machine is modelled by means of a continuous random variable, <math>X</math>, with probability density function</p> |
<p><math>g(x)=\left\{ | <p><math>g(x)=\left\{ | ||
\begin{array}{c} | \begin{array}{c} | ||
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<ol> | <ol> | ||
<li><p>Verify that <math>\int_{0}^{1}g(x)dx=1</math> and calculate <math>\Pr (X\leq 0.5)</math>.</p></li> | <li><p>Verify that <math>\int_{0}^{1}g(x)dx=1</math> and calculate <math>\Pr (X\leq 0.5)</math>.</p></li> | ||
− | <li><p>How much fuel should be supplied each week if the machine is to run out fuel <math>10\%</math> of the time at most? (Note that if <math>s</math> denotes the supply of fuel, then the machine will run out if <math>X>s</math>.)</p></li></ol> | + | <li><p>How much fuel should be supplied each week if the machine is to run out fuel <math>10\%</math> of the time at most? (Note that if <math>s</math> denotes the supply of fuel, then the machine will run out if <math>X>s</math>.) Find the formula you need to solve and find an approximate solution. {approximately 0.54}</p></li></ol> |
</li> | </li> | ||
− | <li><p>The lifetime of a electrical component is measured in <math>100</math>s of hours by a random variable <math>T</math> having the following probability density function</p> | + | <li><p><math>[L2]</math> The lifetime of a electrical component is measured in <math>100</math>s of hours by a random variable <math>T</math> having the following probability density function</p> |
<p><math>f(t)=\left\{ | <p><math>f(t)=\left\{ | ||
\begin{array}{c} | \begin{array}{c} | ||
Line 50: | Line 52: | ||
\right.</math></p></li> | \right.</math></p></li> | ||
<li><p>Show the probability that a component will operate for at least <math>200</math> hours without failure is <math>\Pr (T\geq 2)\cong 0.135</math>.?</p></li> | <li><p>Show the probability that a component will operate for at least <math>200</math> hours without failure is <math>\Pr (T\geq 2)\cong 0.135</math>.?</p></li> | ||
− | <li><p>Three of these electrical components operate independently of one another in a piece of equipment and the equipment fails if ANY ONE of the individual components fail. What is the probability that the equipment will operate for at least <math>200</math> hours without failure? (Use the result in (5.2) in a binomial context).</p></li></ol> | + | <li><p>Three of these electrical components operate independently of one another in a piece of equipment and the equipment fails if ANY ONE of the individual components fail. What is the probability that the equipment will operate for at least <math>200</math> hours without failure? (Use the result in (5.2) in a binomial context). {0.0025}</p></li></ol> |
</li></ol> | </li></ol> | ||
= Footnotes = | = Footnotes = |
Latest revision as of 07:48, 28 August 2014
Probability Exercises
Levels in [square brackets] and solutions in {curly brackets}. Worked solution clips can be found here Q1, Q2, Q3, Q4 and Q5.
[math][L1,L2][/math] In an experiment, if a mouse is administered dosage level [math]A[/math] of a certain (harmless) hormone then there is a [math]0.2[/math] probability that the mouse will show signs of aggression within one minute. For dosage levels [math]B[/math] and [math]C[/math], the probabilities are [math]0.5[/math] and [math]0.8[/math], respectively. Ten mice are given exactly the same dosage level of the hormone and, of these, exactly [math]6[/math] shows signs of aggression within one minute of receiving the dose.
Calculate the probability of this happening for each of the three dosage levels, [math]A,B[/math] and [math]C[/math]. (This is essentially a Binomial random variable problem, so you can check your answers using EXCEL.) {0.0055; 0.2051; 0.0881}
Assuming that each of the three dosage levels was equally likely to have been administered in the first place (with a probability of [math]1/3[/math]), use Bayes’ Theorem to evaluate the likelihood of each of the dosage levels given that [math]6[/math] out of the [math]10[/math] mice were observed to react in this way. Most likely that dosis B was administered.
[math][L1,L2][/math] Let [math]X[/math] be the random variable indicating the number of incoming planes every [math]k[/math] minutes at a large international airport, with probability mass function given by [math]p(x)=\Pr (X=x)=\frac{(0.9k)^{x}}{x!}\exp (-0.9k),\quad x=0,1,2,3,4,..[/math]. Find the probabilities that there will be
exactly [math]9[/math] incoming planes during a period of [math]5[/math] minutes (i.e., find [math]\Pr (X=9)[/math] when [math]k=5)[/math]; { 0.02316}
fewer than [math]5[/math] incoming planes during a period of [math]4[/math] minutes (i.e., find [math]\Pr (X\lt 5)[/math] when [math]k=4)[/math]; { 0.7065}
at least [math]4[/math] incoming planes during an [math]2[/math] minute period (i.e., find [math]\Pr (X\geq 4)[/math] when [math]k=2)[/math]. { 0.1087}
Check all your answers using EXCEL.
[math][L2][/math] The random variable [math]Y[/math] is said to be Geometric if it has probability mass function given by [math]p(y)=\Pr (Y=y)=(1-\theta )\theta ^{y-1},\quad y=1,2,3,...;\quad 0\lt \theta \lt 1[/math]; where [math]\theta [/math] is an unknown ‘parameter’.
Show that the cumulative distribution function can be expressed as
[math]P(y)=\Pr (Y\leq y)=1-\theta ^{y},\quad y=1,2,3,...[/math]
with [math]P(y)=0[/math] for [math]y\leq 0[/math] and [math]P(y)\rightarrow 1[/math] as [math]y\rightarrow \infty. [/math]
(Note that [math]P(y)=p(1)+p(2)+...+p(y)=\sum_{t=1}^{y}p(t)[/math] can be written in longhand as
[math]P(y)=\left( 1-\theta \right) \left( 1+\theta +\theta ^{2}+\theta ^{3}+\ldots +\theta ^{y-1}\right) .[/math]
The term in the second bracket on the right-hand side is the sum of a Geometric Progression.)
[math][L2][/math] The weekly consumption of fuel for a certain machine is modelled by means of a continuous random variable, [math]X[/math], with probability density function
[math]g(x)=\left\{ \begin{array}{c} 3(1-x)^{2},\quad 0\leq x\leq 1, \\ 0,\quad \text{otherwise}.\end{array} \right.[/math]
Consumption, [math]X[/math], is measured in hundreds of gallons per week.
Verify that [math]\int_{0}^{1}g(x)dx=1[/math] and calculate [math]\Pr (X\leq 0.5)[/math].
How much fuel should be supplied each week if the machine is to run out fuel [math]10\%[/math] of the time at most? (Note that if [math]s[/math] denotes the supply of fuel, then the machine will run out if [math]X\gt s[/math].) Find the formula you need to solve and find an approximate solution. {approximately 0.54}
[math][L2][/math] The lifetime of a electrical component is measured in [math]100[/math]s of hours by a random variable [math]T[/math] having the following probability density function
[math]f(t)=\left\{ \begin{array}{c} \exp (-t),\quad t\gt 0, \\ 0,\quad \text{otherwise}.\end{array} \right.[/math]
Show that the cumulative distribution function, [math]F(t)=\Pr (T\leq t)[/math] is given by
[math]F(t)=\left\{ \begin{array}{ll} 1-\exp (-t), & t\gt 0 \\ 0 & t\leq 0.\end{array} \right.[/math]
Show the probability that a component will operate for at least [math]200[/math] hours without failure is [math]\Pr (T\geq 2)\cong 0.135[/math].?
Three of these electrical components operate independently of one another in a piece of equipment and the equipment fails if ANY ONE of the individual components fail. What is the probability that the equipment will operate for at least [math]200[/math] hours without failure? (Use the result in (5.2) in a binomial context). {0.0025}