Probability solutions
This commit is contained in:
+351
-154
@@ -1,147 +1,165 @@
|
||||
<section xml:id="sec-Probability" xmlns:xi="http://www.w3.org/2001/XInclude">
|
||||
<title>Definition of Probability</title>
|
||||
<title>Definition of Probability</title>
|
||||
|
||||
<p>
|
||||
Now that we have the language to refer to outcomes and events of an experiment, we want to start quantifying how likely those outcomes/events are to occur.
|
||||
</p>
|
||||
<p>
|
||||
Now that we have the language to refer to outcomes and events of an
|
||||
experiment, we want to start quantifying how likely those outcomes/events
|
||||
are to occur.
|
||||
</p>
|
||||
|
||||
<definition xml:id="def-probability-distribution">
|
||||
<statement>
|
||||
<p>
|
||||
A <term>probability distribution</term> on a sample space <m>\Omega</m> assigns probabilities to every event, satisfying the following conditions:
|
||||
<ol>
|
||||
<li>
|
||||
<p>
|
||||
<m>\Pr(\Omega) = 1</m>.
|
||||
</p>
|
||||
</li>
|
||||
<definition xml:id="def-probability-distribution">
|
||||
<statement>
|
||||
<p>
|
||||
A <term>probability distribution</term> on a sample space <m>\Omega</m>
|
||||
assigns probabilities to every event, satisfying the following
|
||||
conditions:
|
||||
<ol>
|
||||
<li>
|
||||
<p>
|
||||
<m>\Pr(\Omega) = 1</m>.
|
||||
</p>
|
||||
</li>
|
||||
|
||||
<li>
|
||||
<p>
|
||||
<m>0 \leq \Pr(A) \leq 1</m> for any event <m>A</m>.
|
||||
</p>
|
||||
</li>
|
||||
<li>
|
||||
<p>
|
||||
<m>0 \leq \Pr(A) \leq 1</m> for any event <m>A</m>.
|
||||
</p>
|
||||
</li>
|
||||
|
||||
<li>
|
||||
<p>
|
||||
If <m>A \cap B = \emptyset</m>, then <m>\Pr(A\cup B) = \Pr(A) + \Pr(B)</m>.
|
||||
</p>
|
||||
</li>
|
||||
</ol>
|
||||
</p>
|
||||
</statement>
|
||||
</definition>
|
||||
<li>
|
||||
<p>
|
||||
If <m>A \cap B = \emptyset</m>, then
|
||||
<m>\Pr(A\cup B) = \Pr(A) + \Pr(B)</m>.
|
||||
</p>
|
||||
</li>
|
||||
</ol>
|
||||
</p>
|
||||
</statement>
|
||||
</definition>
|
||||
|
||||
<p>
|
||||
For small probability spaces (i.e., with finitely many outcomes in the sample space), we'll usually assign probabilities to each individual outcome, and perhaps list them in a table.
|
||||
Then, to find the probability of any event, simply add together the probabilities of each outcome in that event.
|
||||
</p>
|
||||
<p>
|
||||
For small probability spaces (i.e., with finitely many outcomes in the
|
||||
sample space), we'll usually assign probabilities to each individual
|
||||
outcome, and perhaps list them in a table.
|
||||
Then, to find the probability of any event, simply add together the
|
||||
probabilities of each outcome in that event.
|
||||
</p>
|
||||
|
||||
<example>
|
||||
<statement>
|
||||
<p>
|
||||
An experiment consists of rolling a standard 6-sided die.
|
||||
The sample space is <m>\Omega = \{1, 2, 3, 4, 5, 6\}</m>.
|
||||
The probability distribution (assuming a fair die) is shown below.
|
||||
</p>
|
||||
<example>
|
||||
<statement>
|
||||
<p>
|
||||
An experiment consists of rolling a standard 6-sided die.
|
||||
The sample space is <m>\Omega = \{1, 2, 3, 4, 5, 6\}</m>.
|
||||
The probability distribution (assuming a fair die) is shown below.
|
||||
</p>
|
||||
|
||||
<table>
|
||||
<title>Distribution for a fair die</title>
|
||||
<table>
|
||||
<title>Distribution for a fair die</title>
|
||||
|
||||
<tabular halign="center">
|
||||
<row bottom="minor">
|
||||
<cell><m>x</m></cell>
|
||||
<cell><m>\Pr(x)</m></cell>
|
||||
</row>
|
||||
<tabular halign="center">
|
||||
<row bottom="minor">
|
||||
<cell><m>x</m></cell>
|
||||
<cell><m>\Pr(x)</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>1</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
<row>
|
||||
<cell>1</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>2</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
<row>
|
||||
<cell>2</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>3</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
<row>
|
||||
<cell>3</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>4</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
<row>
|
||||
<cell>4</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>5</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
<row>
|
||||
<cell>5</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>6</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
</tabular>
|
||||
</table>
|
||||
<row>
|
||||
<cell>6</cell>
|
||||
<cell><m>1/6</m></cell>
|
||||
</row>
|
||||
</tabular>
|
||||
</table>
|
||||
|
||||
<p>
|
||||
One possible event is <m>A = \{2, 4, 6\}</m>, i.e., the event that the result of the roll is even.
|
||||
The probability of <m>A</m> is:
|
||||
<md>
|
||||
<mrow> \Pr(A) = \Pr(2) + \Pr(4) + \Pr(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} </mrow>
|
||||
</md>
|
||||
</p>
|
||||
</statement>
|
||||
</example>
|
||||
<p>
|
||||
One possible event is <m>A = \{2, 4, 6\}</m>, i.e., the event that the
|
||||
result of the roll is even.
|
||||
The probability of <m>A</m> is:
|
||||
<md>
|
||||
<mrow> \Pr(A) = \Pr(2) + \Pr(4) + \Pr(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} </mrow>
|
||||
</md>
|
||||
</p>
|
||||
</statement>
|
||||
</example>
|
||||
|
||||
<definition xml:id="def-discrete-uniform-distribution">
|
||||
<statement>
|
||||
<p>
|
||||
Let <m>\Omega = \{x_1, x_2, \dotsc, x_n\}</m> be a sample space and <m>A\subset \Omega</m> an event.
|
||||
We refer to the distribution in which <m>\Pr(x_i) = \frac{1}{n}</m> for all <m>i</m> as the <term>uniform distribution</term>.
|
||||
In this case, it follows that <m>\Pr(A) = \frac{|A|}{|\Omega|}</m>.
|
||||
</p>
|
||||
</statement>
|
||||
</definition>
|
||||
<definition xml:id="def-discrete-uniform-distribution">
|
||||
<statement>
|
||||
<p>
|
||||
Let <m>\Omega = \{x_1, x_2, \dotsc, x_n\}</m> be a sample space and
|
||||
<m>A\subset \Omega</m> an event.
|
||||
We refer to the distribution in which <m>\Pr(x_i) = \frac{1}{n}</m> for
|
||||
all <m>i</m> as the <term>uniform distribution</term>.
|
||||
In this case, it follows that <m>\Pr(A) = \frac{|A|}{|\Omega|}</m>.
|
||||
</p>
|
||||
</statement>
|
||||
</definition>
|
||||
|
||||
<p>
|
||||
When we talk about a fair coin flip or a fair die roll, the word "fair" is indicating a uniform distribution.
|
||||
However, don't make the mistake of assuming that all distributions are uniform by default.
|
||||
</p>
|
||||
<p>
|
||||
When we talk about a fair coin flip or a fair die roll, the word "fair" is
|
||||
indicating a uniform distribution.
|
||||
However, don't make the mistake of assuming that all distributions are
|
||||
uniform by default.
|
||||
</p>
|
||||
|
||||
<example>
|
||||
<statement>
|
||||
<p>
|
||||
A person picks a random number from 1 to 10.
|
||||
What is the probability that they picked 3?
|
||||
</p>
|
||||
</statement>
|
||||
<example>
|
||||
<statement>
|
||||
<p>
|
||||
A person picks a random number from 1 to 10.
|
||||
What is the probability that they picked 3?
|
||||
</p>
|
||||
</statement>
|
||||
|
||||
<answer>
|
||||
<p>
|
||||
Without assuming the distribution is fair (i.e., that each value <m>1, 2, \dotsc, 10</m>) has probability <m>1/10</m> of occurring), we don't have enough information to answer this question.
|
||||
</p>
|
||||
<answer>
|
||||
<p>
|
||||
Without assuming the distribution is fair (i.e., that each value
|
||||
<m>1, 2, \dotsc, 10</m> ) has probability <m>1/10</m> of occurring), we
|
||||
don't have enough information to answer this question.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
In fact, the situation is even more vague than that: the sample space itself is unclear.
|
||||
Are we only allowed to pick integer values? What about fractions like <m>7/2</m>? What about irrational numbers like <m>\pi</m>?
|
||||
</p>
|
||||
</answer>
|
||||
</example>
|
||||
|
||||
<exercises xml:id="exercises-Probability">
|
||||
<p>
|
||||
In fact, the situation is even more vague than that: the sample space
|
||||
itself is unclear.
|
||||
Are we only allowed to pick integer values?
|
||||
What about fractions like <m>7/2</m>?
|
||||
What about irrational numbers like <m>\pi</m>?
|
||||
</p>
|
||||
</answer>
|
||||
</example>
|
||||
<exercises xml:id="exercises-Probability">
|
||||
<exercise>
|
||||
<statement>
|
||||
<p>
|
||||
Consider the sample space <m>\Omega = \{1, 2, 3, 4, 5, 6, 7, 8\}</m> with probability distribution below.
|
||||
Calculate the probabilities of <m>A = \{1, 3, 7, 8\}</m>, <m>B = \{2, 3, 6, 7\}</m>, <m>A\cup B</m>, and <m>A \cap B</m>.
|
||||
Consider the sample space <m>\Omega = \{1, 2, 3, 4, 5, 6, 7, 8\}</m>
|
||||
with probability distribution below.
|
||||
Calculate the probabilities of <m>A = \{1, 3, 7, 8\}</m>,
|
||||
<m>B = \{2, 3, 6, 7\}</m>, <m>A\cup B</m>, and <m>A \cap B</m>.
|
||||
</p>
|
||||
|
||||
<table>
|
||||
<title></title>
|
||||
|
||||
<title/>
|
||||
<tabular halign="center">
|
||||
<row bottom="minor">
|
||||
<cell><m>x</m></cell>
|
||||
@@ -196,11 +214,41 @@
|
||||
</table>
|
||||
</statement>
|
||||
|
||||
<hint>
|
||||
<p>
|
||||
Remember that, to calculate the probability of an event, you should
|
||||
add up the probabilities of each outcome in the event.
|
||||
</p>
|
||||
</hint>
|
||||
|
||||
<answer>
|
||||
<p>
|
||||
<m>\Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25.</m>
|
||||
</p>
|
||||
</answer>
|
||||
|
||||
<solution>
|
||||
<p>
|
||||
<md>
|
||||
<mrow> \Pr(A) \amp = \Pr(1) + \Pr(3) + \Pr(7) + \Pr(8) </mrow>
|
||||
<mrow> \amp = 0.1 + 0.2 + 0.05 + 0.1 </mrow>
|
||||
<mrow> \amp = 0.45 </mrow>
|
||||
<mrow> \Pr(B) \amp = \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) </mrow>
|
||||
<mrow> \amp = 0.05 + 0.2 + 0.1 + 0.05 </mrow>
|
||||
<mrow> \amp = 0.4 </mrow>
|
||||
</md>
|
||||
The other events are <m>A\cup B = \{1, 2, 3, 6, 7, 8\}</m> and
|
||||
<m>A\cap B = \{3, 7\}</m>, so:
|
||||
<md>
|
||||
<mrow> \Pr(A \cup B) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) + \Pr(8) </mrow>
|
||||
<mrow> \amp = 0.1 + 0.05 + 0.2 + 0.1 + 0.05 + 0.1 </mrow>
|
||||
<mrow> \amp = 0.6 </mrow>
|
||||
<mrow> \Pr(A \cap B) \amp = \Pr(3) + \Pr(7) </mrow>
|
||||
<mrow> \amp = 0.2 + 0.05 </mrow>
|
||||
<mrow> \amp = 0.25 </mrow>
|
||||
</md>
|
||||
</p>
|
||||
</solution>
|
||||
</exercise>
|
||||
|
||||
<exercise>
|
||||
@@ -208,11 +256,11 @@
|
||||
<p>
|
||||
Suppose we flip a coin two times.
|
||||
Answer the questions below.
|
||||
What about three flips? What about four flips?
|
||||
What about three flips?
|
||||
What about four flips?
|
||||
</p>
|
||||
</introduction>
|
||||
|
||||
|
||||
<task>
|
||||
<statement>
|
||||
<p>
|
||||
@@ -227,11 +275,11 @@
|
||||
</answer>
|
||||
</task>
|
||||
|
||||
|
||||
<task>
|
||||
<statement>
|
||||
<p>
|
||||
Make a probability distribution table for <m>\Omega</m> assuming the coin is fair.
|
||||
Make a probability distribution table for <m>\Omega</m> assuming the
|
||||
coin is fair.
|
||||
</p>
|
||||
</statement>
|
||||
|
||||
@@ -269,11 +317,11 @@
|
||||
</answer>
|
||||
</task>
|
||||
|
||||
|
||||
<task>
|
||||
<statement>
|
||||
<p>
|
||||
Make a probability distribution table assuming the coin comes up heads with probability 0.3.
|
||||
Make a probability distribution table assuming the coin comes up
|
||||
heads with probability 0.3.
|
||||
</p>
|
||||
</statement>
|
||||
|
||||
@@ -315,12 +363,11 @@
|
||||
<exercise>
|
||||
<introduction>
|
||||
<p>
|
||||
Suppose we roll a die two times.
|
||||
Suppose we roll a fair 6-sided die two times.
|
||||
Answer the questions below.
|
||||
</p>
|
||||
</introduction>
|
||||
|
||||
|
||||
<task>
|
||||
<statement>
|
||||
<p>
|
||||
@@ -342,32 +389,40 @@
|
||||
</answer>
|
||||
</task>
|
||||
|
||||
|
||||
<task>
|
||||
<statement>
|
||||
<p>
|
||||
Make a probability distribution table for <m>\Omega</m> assuming the die is fair.
|
||||
Make a probability distribution table for <m>\Omega</m> assuming the
|
||||
die is fair.
|
||||
</p>
|
||||
</statement>
|
||||
|
||||
<answer>
|
||||
<p>
|
||||
We'll avoid an overly large table and note that, since the die is fair, every outcome is equally likely.
|
||||
Therefore, <m>\Pr(x) = \frac{1}{36}</m> for every <m>x\in \Omega</m>.
|
||||
We'll avoid an overly large table and note that, since the die is
|
||||
fair, every outcome is equally likely.
|
||||
Therefore, <m>\Pr(x) = \frac{1}{36}</m> for every
|
||||
<m>x\in \Omega</m>.
|
||||
</p>
|
||||
</answer>
|
||||
</task>
|
||||
|
||||
|
||||
<task>
|
||||
<statement>
|
||||
<p>
|
||||
Let <m>A</m> be the event that the second roll is higher than the first, and let <m>B</m> be the event that the first roll is even.
|
||||
Let <m>A</m> be the event that the second roll is higher than the
|
||||
first, and let <m>B</m> be the event that the first roll is even.
|
||||
Find <m>\Pr(A), \Pr(B)</m>, and <m>\Pr(A \cap B)</m>.
|
||||
</p>
|
||||
</statement>
|
||||
|
||||
<answer>
|
||||
<p>
|
||||
<m>\Pr(A) = \frac{5}{12}, \Pr(B) = \frac{1}{2}, \Pr(A \cap B) = \frac{1}{6}</m>.
|
||||
</p>
|
||||
</answer>
|
||||
|
||||
<solution>
|
||||
<p>
|
||||
<md>
|
||||
<mrow> A = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), </mrow>
|
||||
@@ -381,18 +436,23 @@
|
||||
<mrow> A \cap B = \{ \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
|
||||
<mrow> \amp (4, 5), (4, 6)\} </mrow>
|
||||
</md>
|
||||
Therefore <m>\Pr(A) = \frac{15}{36} = \frac{5}{12}, \Pr(B) = \frac{18}{36} = \frac{1}{2}, \Pr(A \cap B) = \frac{6}{36} = \frac{1}{6}.</m>
|
||||
Therefore
|
||||
<m>\Pr(A) = \frac{15}{36} = \frac{5}{12}, \Pr(B) = \frac{18}{36} = \frac{1}{2}, \Pr(A \cap B) = \frac{6}{36} = \frac{1}{6}.</m>
|
||||
</p>
|
||||
</answer>
|
||||
</solution>
|
||||
</task>
|
||||
</exercise>
|
||||
|
||||
<exercise>
|
||||
<statement>
|
||||
<p>
|
||||
Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but the die is not fair.
|
||||
Instead, the probabilities scale by the same amount as the face values.
|
||||
For example, a result of 4 is twice as likely as a result of 2, since 4 is twice as large as 2; a result of 6 is six times more likely than a result of 1; and so on.
|
||||
Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but
|
||||
the die is not fair.
|
||||
Instead, the probabilities scale by the same amount as the face
|
||||
values.
|
||||
For example, a result of 4 is twice as likely as a result of 2, since
|
||||
4 is twice as large as 2; a result of 6 is six times more likely than
|
||||
a result of 1; and so on.
|
||||
Write a probability distribution table for this die.
|
||||
</p>
|
||||
</statement>
|
||||
@@ -439,13 +499,74 @@
|
||||
</tabular>
|
||||
</table>
|
||||
</answer>
|
||||
|
||||
<solution>
|
||||
<p>
|
||||
Let <m>\Pr(1) = x</m>.
|
||||
Then <m>\Pr(2) = 2 \Pr(1) = 2x</m>, and <m>\Pr(3) = 3\Pr(1) = 3x</m>,
|
||||
and so on.
|
||||
So the total probability in the space is:
|
||||
<md>
|
||||
<mrow> Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) </mrow>
|
||||
<mrow> \amp = x + 2x + 3x + 4x + 5x + 6x </mrow>
|
||||
<mrow> \amp = 21x </mrow>
|
||||
</md>
|
||||
Since the total probability must add up to 1, we have <m>1 = 21x</m>,
|
||||
so <m>x = \frac{1}{21}</m>, and we can calculate the rest of the
|
||||
probabilities from there.
|
||||
</p>
|
||||
|
||||
<table>
|
||||
<title>Probability Distribution for a Linearly Scaled Die</title>
|
||||
|
||||
<tabular halign="center">
|
||||
<row bottom="minor">
|
||||
<cell><m>x</m></cell>
|
||||
<cell><m>\Pr(x)</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>1</cell>
|
||||
<cell><m>1/21</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>2</cell>
|
||||
<cell><m>2/21</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>3</cell>
|
||||
<cell><m>3/21</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>4</cell>
|
||||
<cell><m>4/21</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>5</cell>
|
||||
<cell><m>5/21</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>6</cell>
|
||||
<cell><m>6/21</m></cell>
|
||||
</row>
|
||||
</tabular>
|
||||
</table>
|
||||
</solution>
|
||||
</exercise>
|
||||
|
||||
<exercise>
|
||||
<statement>
|
||||
<p>
|
||||
Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but the die is not fair.
|
||||
Instead, each even value has an equal probability, each odd value has an equal probability, and the even values are each twice as likely as the odd values to appear on a roll.
|
||||
Suppose a die has the values <m>1, 2, 3, 4, 5, 6</m> on the faces, but
|
||||
the die is not fair.
|
||||
Instead, each even value has an equal probability, each odd value has
|
||||
an equal probability, and the even values are each twice as likely as
|
||||
the odd values to appear on a roll.
|
||||
Write a probability distribution table for this die.
|
||||
</p>
|
||||
</statement>
|
||||
@@ -492,36 +613,112 @@
|
||||
</tabular>
|
||||
</table>
|
||||
</answer>
|
||||
|
||||
<solution>
|
||||
<p>
|
||||
Let <m>\Pr(1) = x</m>.
|
||||
Then <m>\Pr(3) = x</m> and <m>\Pr(5) = x</m>, since all odd rolls must
|
||||
have the same probability.
|
||||
The even rolls must have twice the probability, so
|
||||
<m>\Pr(2) = \Pr(4) = \Pr(6) = 2x</m>.
|
||||
Now:
|
||||
<md>
|
||||
<mrow> Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) </mrow>
|
||||
<mrow> \amp = x + 2x + x + 2x + x + 2x </mrow>
|
||||
<mrow> \amp = 9x </mrow>
|
||||
</md>
|
||||
Since the total probability must add up to 1, we have <m>1 = 9x</m>,
|
||||
so <m>x = \frac{1}{9}</m>, and we can calculate the rest of the
|
||||
probabilities from there.
|
||||
</p>
|
||||
|
||||
<table>
|
||||
<title>Probability Distribution for an Even-biased Die</title>
|
||||
|
||||
<tabular halign="center">
|
||||
<row bottom="minor">
|
||||
<cell><m>x</m></cell>
|
||||
<cell><m>\Pr(x)</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>1</cell>
|
||||
<cell><m>1/9</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>2</cell>
|
||||
<cell><m>2/9</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>3</cell>
|
||||
<cell><m>1/9</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>4</cell>
|
||||
<cell><m>2/9</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>5</cell>
|
||||
<cell><m>1/9</m></cell>
|
||||
</row>
|
||||
|
||||
<row>
|
||||
<cell>6</cell>
|
||||
<cell><m>2/9</m></cell>
|
||||
</row>
|
||||
</tabular>
|
||||
</table>
|
||||
</solution>
|
||||
</exercise>
|
||||
|
||||
<exercise>
|
||||
<statement>
|
||||
<p>
|
||||
A toxin molecule inside a cell has a 0.3 probability of leaving the cell during a 1-minute period.
|
||||
For each value of <m>n = 1, 2, 3, \dotsc</m>, find the probability of the toxin molecule leaving the cell during the <m>n</m>th minute.
|
||||
What is the probability of the molecule leaving the cell during the first 3 minutes?
|
||||
A toxin molecule inside a cell has a 0.3 probability of leaving the
|
||||
cell during a 1-minute period.
|
||||
For each value of <m>n = 1, 2, 3, \dotsc</m>, find the probability of
|
||||
the toxin molecule leaving the cell during the <m>n</m> th minute.
|
||||
What is the probability of the molecule leaving the cell during the
|
||||
first 3 minutes?
|
||||
</p>
|
||||
</statement>
|
||||
|
||||
<answer>
|
||||
<p>
|
||||
For short, write <m>\Pr(n)</m> to mean the probability of the toxin molecule leaving during the <m>n</m>th minute.
|
||||
Then <m>\Pr(n) = (0.7)^{n - 1} (0.3).</m>
|
||||
Write <m>T</m> for the minute that the toxin molecule leaves the cell.
|
||||
Then <m>\Pr(T = n) = (0.7)^{n - 1} (0.3)</m>, and <m>\Pr(T \leq 3) = 0.657</m>.
|
||||
</p>
|
||||
</answer>
|
||||
|
||||
<solution>
|
||||
<p>
|
||||
Write <m>T</m> for the minute that the toxin molecule leaves the cell.
|
||||
We're told that the toxin molecule has probability <m>0.3</m> of leaving during each minute.
|
||||
Therefore, the probability of remaining during a particular minute is <m>1 - 0.3 = 0.7</m>.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
The probability of leaving during the first 3 minutes is <m>\Pr(1) + \Pr(2) + \Pr(3) = 0.657.</m>
|
||||
For the toxin molecule to leave the cell during minute <m>n</m>, it must remain for minutes <m>1, 2, \dotsc, n - 1</m>, and then leave during minute <m>n</m>.
|
||||
The probability to remain each minute is <m>0.7</m>, so we must multiply <m>n - 1</m> copies of <m>0.7</m>.
|
||||
Then the probability of leaving is <m>0.3</m>, so we multiply by a factor of <m>0.3</m>, yielding the formula:
|
||||
<md>
|
||||
<mrow> \Pr(T = n) = (0.7)^{n - 1}(0.3). </mrow>
|
||||
</md>
|
||||
</p>
|
||||
</answer>
|
||||
</exercise>
|
||||
<!--
|
||||
<exercise>
|
||||
<statement>
|
||||
|
||||
<p>
|
||||
Each of 10 toxin molecules inside a cell has a 0.3 probability of leaving the cell during a 1-minute period.
|
||||
For each value of <m>n = 1, 2, 3, \dotsc</m>, and for each value of <m>0\leq k \leq n</m>, find the probability that exactly <m>k</m> toxin molecules remain in the cell after the <m>n</m>th minute.
|
||||
Then, the probaiblity of the molecule leaving within the first three minutes will be:
|
||||
<md>
|
||||
<mrow> \Pr(T \leq 3) \amp = \Pr(T = 1) + \Pr(T = 2) + \Pr(T = 3) </mrow>
|
||||
<mrow> \amp = 0.3 + (0.7)(0.3) + (0.7)^2(0.3) </mrow>
|
||||
<mrow> \amp = 0.657. </mrow>
|
||||
</md>
|
||||
</p>
|
||||
</statement>
|
||||
</solution>
|
||||
</exercise>
|
||||
needs binomial coefficients --> </exercises>
|
||||
</section>
|
||||
</exercises>
|
||||
</section>
|
||||
Reference in New Issue
Block a user