Introductory Probability (2 Viewers)

leehuan · May 9, 2016

Feeling like I may have unnecessarily overcomplicated this question just to get the right answer. Would appreciate advice on if there is a shorter method. (I think I am required to use set notation.)

$Q: A brewery brews one tpe of beer which is marketed under three different brands. In a survey of 150 first year students, 58 drink at least brand A, 49 drink at least brand B and 57 drink at least brand C. \\ 14 drink brand A and brand C\\ 13 drink brand A and brand B\\ and 17 drink both brand B and brand C. \\ 4 students drink all 3 brands.\\ How many students drink none of these brands?$

Answer:
$P\left( { \left( A\cup B\cup C \right) }^{ C } \right) \\ =1-P\left( A\cup B\cup C \right) \\ =1-\left[ P\left( A\cup B \right) +P\left( C \right) -P\left( \left( A\cup B \right) \cap C \right) \right] \\ =1-\left[ \left( P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \right) +P\left( C \right) -P\left( \left( A\cap C \right) \cup \left( B\cap C \right) \right) \right] \\ =1-\left[ \frac { 58+49-13+57 }{ 150 } -\left( P\left( A\cap C \right) +P\left( B\cap C \right) -P\left( \left( A\cap C \right) \cap \left( B\cap C \right) \right) \right) \right] \\ =1-\left[ \frac { 151 }{ 150 } -\left( \frac { 14+17 }{ 150 } -P\left( A\cap B\cap C \right) \right) \right] \\ =1-\left( \frac { 120 }{ 150 } +\frac { 4 }{ 150 } \right) \\ =\frac { 13 }{ 75 }$
$Hence 26 people$

InteGrand · May 9, 2016

leehuan said:
Feeling like I may have unnecessarily overcomplicated this question just to get the right answer. Would appreciate advice on if there is a shorter method. (I think I am required to use set notation.)

$Q: A brewery brews one tpe of beer which is marketed under three different brands. In a survey of 150 first year students, 58 drink at least brand A, 49 drink at least brand B and 57 drink at least brand C. \\ 14 drink brand A and brand C\\ 13 drink brand A and brand B\\ and 17 drink both brand B and brand C. \\ 4 students drink all 3 brands.\\ How many students drink none of these brands?$

Answer:
$P\left( { \left( A\cup B\cup C \right) }^{ C } \right) \\ =1-P\left( A\cup B\cup C \right) \\ =1-\left[ P\left( A\cup B \right) +P\left( C \right) -P\left( \left( A\cup B \right) \cap C \right) \right] \\ =1-\left[ \left( P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \right) +P\left( C \right) -P\left( \left( A\cap C \right) \cup \left( B\cap C \right) \right) \right] \\ =1-\left[ \frac { 58+49-13+57 }{ 150 } -\left( P\left( A\cap C \right) +P\left( B\cap C \right) -P\left( \left( A\cap C \right) \cap \left( B\cap C \right) \right) \right) \right] \\ =1-\left[ \frac { 151 }{ 150 } -\left( \frac { 14+17 }{ 150 } -P\left( A\cap B\cap C \right) \right) \right] \\ =1-\left( \frac { 120 }{ 150 } +\frac { 4 }{ 150 } \right) \\ =\frac { 13 }{ 75 }$
$Hence 26 people$

You can also try a Venn Diagram approach. (Since there are only three sets here, it's easy to draw the diagram. If there are more, it gets harder to draw.)

Also you can use the inclusion-exclusion formula early on (I think you essentially derived it). For three sets, it is

$|A \cup B \cup C| = |A|+|B|+|C| -|A\cap B| - |B\cap C| - |C \cap A | + |A \cap B \cap C|$ .

leehuan · May 10, 2016

$Urn 1 contains 2 red balls and 3 black balls. Urn 2 contains 4 red balls and 5 black balls.$\\ $b) Suppose a ball is drawn at random from urn 1 and placed into urn 2. If a ball is then drawn at random from the 10 balls in urn 2, what is the probability that it is red$

Conditional probaility, but I am clueless here because I have no idea how to combine the two seperate cases of a red ball being transferred, or a black ball being transferred

Probably not useful part a) P(Red (without moving balls)) = 19/45

InteGrand · May 10, 2016

leehuan said:
$Urn 1 contains 2 red balls and 3 black balls. Urn 2 contains 4 red balls and 5 black balls.$\\ $b) Suppose a ball is drawn at random from urn 1 and placed into urn 2. If a ball is then drawn at random from the 10 balls in urn 2, what is the probability that it is red$

Conditional probaility, but I am clueless here because I have no idea how to combine the two seperate cases of a red ball being transferred, or a black ball being transferred

Probably not useful part a) P(Red (without moving balls)) = 19/45

Let B1 be the event that a black ball was transferred from Urn 1 to Urn 2. Let R1 be a similar event for red (red transferred from Urn 1 to Urn 2).

Let B2 and R2 respectively be the events that black and red are drawn from Urn 2 after the transfer from Urn 1 has happened.

We want to find Pr(R2).

By the law of total probability,

Pr(R2) = Pr(R2 | B1)*Pr(B1) + Pr(R2 | R1)*Pr(R1) (this is valid since R1 and B1 are complementary events).

Now, Pr(R2 | B1) = 4/10, Pr(B1) = 3/5, Pr(R2 | R1) = 5/10 and Pr(R1) = 2/5 (for the conditional probabilities, we just find probabilities assuming we know which ball got transferred. So for example, Pr(R2 | B1) = 4/10, because given we know a black ball was transferred from Urn 1 to Urn 2, there'll be 10 balls in Urn 2 and 4 of them red, so 4/10).

Plugging these into the above equation yields the answer.

davidgoes4wce · May 10, 2016

leehuan said:
$Urn 1 contains 2 red balls and 3 black balls. Urn 2 contains 4 red balls and 5 black balls.$\\ $b) Suppose a ball is drawn at random from urn 1 and placed into urn 2. If a ball is then drawn at random from the 10 balls in urn 2, what is the probability that it is red$

Conditional probaility, but I am clueless here because I have no idea how to combine the two seperate cases of a red ball being transferred, or a black ball being transferred

Probably not useful part a) P(Red (without moving balls)) = 19/45

This is my way of doing it, I used the tree diagrams.

davidgoes4wce · May 10, 2016

Sorry for the camera quality, I just degraded down to an Iphone 4 because I lost my Iphone 6.

davidgoes4wce · May 10, 2016

leehuan said:
$Urn 1 contains 2 red balls and 3 black balls. Urn 2 contains 4 red balls and 5 black balls.$\\ $b) Suppose a ball is drawn at random from urn 1 and placed into urn 2. If a ball is then drawn at random from the 10 balls in urn 2, what is the probability that it is red$

Conditional probaility, but I am clueless here because I have no idea how to combine the two seperate cases of a red ball being transferred, or a black ball being transferred

Probably not useful part a) P(Red (without moving balls)) = 19/45

Do you mind if you posted the question for Part (a) up........?

davidgoes4wce · May 10, 2016

leehuan · May 10, 2016

Part a) was just

"If an urn is randomly selected and a ball drawn at random from it. what is the probability that it is red?"

P(Red) = P(Red n Urn 1) + P(Red n Urn 2) = rewrite as conditional probability * P(Urn) = (2/5)(1/2) + (4/9)(1/2)

leehuan · May 10, 2016

I don't know what I'm finding. (I think I'm just having hard times today trying to understand the question.)

$Employment data at a large company reveal that 72 percent of the workers are married, that 44 percent are university graduates and that half of the university graduates are married. What is the probability that a randomly chosen worker\\ b) is married but not a university graduate?$

Deduced earlier:
P(M)=72/100, P(U)=44/100
P(M|U)=1/2
P(MnU) = 1/2 * 44/100 = 22/100

P(M^CnU^{C/sup])=3/50 from part (a)

The answer is also 1/2 which made me contemplate P(M|U) + P(M|U^C) = 1 may be how to go about it but I can't justify why}

davidgoes4wce · May 10, 2016

leehuan said:
I don't know what I'm finding. (I think I'm just having hard times today trying to understand the question.)

$Employment data at a large company reveal that 72 percent of the workers are married, that 44 percent are university graduates and that half of the university graduates are married. What is the probability that a randomly chosen worker\\ b) is married but not a university graduate?$

Deduced earlier:
P(M)=72/100, P(U)=44/100
P(M|U)=1/2
P(MnU) = 1/2 * 44/100 = 22/100

P(M^CnU^{C/sup])=3/50 from part (a)

The answer is also 1/2 which made me contemplate P(M|U) + P(M|U^C) = 1 may be how to go about it but I can't justify why}

^{I just simply look at the intersection of Married and not a University graduate region and divide it by the total population. Here is my diagram below:}

davidgoes4wce · May 10, 2016

davidgoes4wce · May 10, 2016

With any conditional probability question, you are technically reducing the denominator numeral. (reducing the population)

leehuan · May 10, 2016

At the start I actually thought I was finding P(MnU^C) but then I realised I had no idea how to actually get there. Now I realise it's the simple P(M) - P(MnU)

Reckon I should just keep drawing Venn diagrams till this stuff becomes intuitive? Lol

seanieg89 · May 10, 2016

leehuan said:
At the start I actually thought I was finding P(MnU^C) but then I realised I had no idea how to actually get there. Now I realise it's the simple P(M) - P(MnU)

Reckon I should just keep drawing Venn diagrams till this stuff becomes intuitive? Lol

Don't rely too much on them, unless you have a handy supply of higher dimensional paper for when there are more than three sets.

Way more important to just internalise how things like taking complements interacts with unions and intersections of sets (from a mathematical point of view outside of combinatorics as well you absolutely need to know this).

davidgoes4wce · May 10, 2016

Tree diagrams, Venn Diagrams and Tables are the best way to learn probability. (I've always been a visualiser, some of the students who I have tutored at unis like it when I present them with diagrams)

I thought it was covered pretty rigorously in HSC Maths Extension 1 and from what I gather in more advanced levels in Extension 2.

InteGrand · May 10, 2016

leehuan · May 10, 2016

InteGrand said:
$\noindent A useful probability identity is $\Pr \left(A\right) = \Pr \left(A \cap B\right) + \Pr \left(A \cap B^c\right)$ (follows from the Law of Total Probability; using set difference notation, can also be written as $\Pr \left(A\right) =\Pr \left(A \cap B\right) + \Pr \left(A - B\right)$). Rearranging this yields $\Pr \left(A \cap B^{c}\right) = \Pr \left(A\right) - \Pr \left(A \cap B\right)$. This identity is the one that was useful for the married / undergraduate problem above.$

You could say this is exactly what I needed at the time

leehuan · May 10, 2016

Internet is too hard to understand. So just a quick startup on the inductive step please?

$$Establish, using mathematical induction, Bonferoni's inequality: $\\ P\left(A_1\cap A_2\cap \dots \cap A_n \right) \ge 1-\left[P(A_1^C)+P(A_2^C)+\dots+P(A_n^C)\right]$

Because I only know how to split up AuC, not AnC

Paradoxica · May 10, 2016

seanieg89 said:
Don't rely too much on them, unless you have a handy supply of higher dimensional paper for when there are more than three sets.

Way more important to just internalise how things like taking complements interacts with unions and intersections of sets (from a mathematical point of view outside of combinatorics as well you absolutely need to know this).

I have a method that can enclose any number of sets...

It simply requires a sufficiently large sheet of paper.

Introductory Probability (2 Viewers)

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