Here’s your next reading assignment. Read Section 2.3 (omitting 2.3.7 for now) in your textbook and answer the following questions by 8 a.m., Wednesday, February 1st. Be sure to login to the blog before leaving your answers in the comment section below.
- Suppose you select a random member of the Vanderbilt class of 2011. Let A be the event that the student was an engineering major, and let B be the event that the student took calculus while in college. Which is greater, P(A | B) or P(B | A)? Why?
- Suppose a factory has two production lines, A and B. Line A produces 40% of the factory’s output, and Line B produces the rest. Of the parts made on Line A, 4% are defective. Of the parts made on Line B, 3% are defective. Suppose a part is picked at random from inventory and found to be defective. What’s the probability it came from line A?
- Suppose you draw a card from a standard 52-card deck at random, then draw a second card at random from the remaining 51 cards. Use the General Multiplication Rule to find the probability that both cards are hearts.
- What’s one question you have about the reading?
1. The P(B | A) is higher. All engineering students have to take calculus (I’m pretty sure), so if you know that a student is an engineering you can basically say with certainty that they’ve taken calculus or will take it while in college. The P (A | B) must be less because not everyone who takes calculus is necessarily an engineering major. Math majors and certain science majors like physics or neuroscience also require students to take calculus but aren’t part of the School of Engineering.
2. Using a tree diagram, the probability comes out to be about 47%.
3. 5.88%.
4. I know it wasn’t required reading, but I don’t really understand how using Bayes’ Theorem to solve problems is different from solving problems like #2. It would be nice to work a lot of problems dealing with decks of cards because I remember messing up those the most in high school.
1. The first probability would be the probability that the student is an engineering major given they have taken calculus. The second would be the probability the student took calculus given they were an engineering majors. This second probability is more likely given that all engineering students must take calculus and the first is less likely as some students have taken calculus but are not engineering majors
2. I’m not very sure on this one but I am guessing 0.04
3. 1/17
4. I am a little confused on the General Multiplication Rule and how to decide the probability of A given B in some of the examples.
1) P(B|A) is greater than P(A|B) since P(B|A) is the probability that a student took a calculus class given that the student was an engineering major. Since (almost) all engineering majors take calculus, this value will be close to 1. P(A|B) is the probability that the student is a engineering major given that the student took a calculus class. Since calculus classes are taken by math, physics, chemistry, and several other majors, this probability would not be as high as P(B|A).
2) P(A|defective)=P(A and defective)/P(defective) = (0.4*0.04)/(0.4*0.04+0.6*0.03) = 0.4706
3) P(2 Hearts)=13/52 * 12/51 = (1/4)*(12/51)=12/204=3/51=0.0588
4) When do we use tree diagrams as opposed to Venn diagrams?
1) P(B|A) would be greater. P(B|A) = P(B and A) / P(A) while P(A|B) = P(A and B) /
P(B). P(B) is greater than P(A) because there are more students taking calculus than there are engineers, so when we divide by P(B), we’ll get a smaller proportion. But when we divide by P(A), we’ll get a greater proportion because P(A) is smaller.
2) 0.47
3) 0.06 (I am sure that’s wrong)
4) So from the reading from exercise 2.58 part c, I came to the conclusion that an easy way to check if 2 events are independent is to see if P(A|B) = P(A). Is it correct?
1. P(B|A) would be bigger b/c in either case the numerator is P(A) & P(B) but the denominator is the second event, in this case A, which we want to be small to get a bigger fraction. P(A) would be smaller b/c # engineering students < # students who took calculus.
2. P(Being from A | being defective) = P(Being from A & it's defective) / P(Being defective) = (.4 * .04) / .07 = .23
3. P(Card1=heart & Card2=heart) = P(Card1=heart | Card2=heart) * P(Card2=heart) = 12/52 * 13/52 = .058
4. I don't think I got 2 right.
1. P(B|A) is greater because, if looking at all Vanderbilt engineering students for class of 2011, there is a 100% probability that they took some sort of calculus in college. Conversely, if looking at all students of class of 2011 that took calculus, only a portion would be engineering students, as some could be math, physics, biology, etc majors.
2. P(line A given defective) = 47.1%
3. P(hearts52 and hearts51) = P(hearts51 given hearts52) * P(hearts52) = (12/51)*(13/52) = 0.235*0.25 = 0.0588 = 5.9%
4. A walkthrough of these methods would be great.
1) P(A|B), because it is more likely for an engineering major to take calculus in college since it is the essential requirements.
2) 0.16
3)0.25(12/51)=0.06
4)Why the general multiplication rule expressed in that way? Wouldn’t P(A|B)*P(B) just equal P(A&B)
1) P(B|A) would be greater since the percentage of engineering students that take calculus is much higher than the percentage of students in engineering that took calculus. Since most, if not all, engineers are required to take calculus, and other majors can take it but are not required to, P(B|A) must be greater.
2) Given that P(A | defect) = .04 * .4 = .016 and P(B|defect) = .03 * .6 = .018, we can find the likelyhood that the product was from line A by dividing the probability of P(A|defect) by the total probability that the product had a defect. .016/(.016+.018) = .47
3) (13/52) * (12/51) = .05882
4) My biggest difficulty with this section is determining when you need to multiply the given percentages by other percentages in the tree diagrams and when the information they give you is sufficient.
1. Hard to tell without numerical information. However, since pretty much all engineering require at least Calc III and AP testing only covers through Calc II, P(B|A) would probably be larger. In contrast, Calc isn’t required in all majors, and many non-engineering students may try to avoid a Calculus class.
2. A: .4*.4=.16. B: .6*.3=.18 Sum is .34.
.16/.34 = 47% chance it came from A
3. 13/52 * 12/51 = 0.0588
4. What’s one question you have about the reading?
None really–pretty straightforward
1) P(B | A) > P(A | B) because all engineering majors (event A) at Vanderbilt are required to take calculus, but not all calculus students (event B) are necessarily engineering majors. Thus, the probability of B given A is higher than the probability of A if given B.
2) Take a factory output of 1000 parts. Of those parts, 400 would be produced on line A and 600 on line B. Of those 400 on line A, 16 would be defective. Of the 600 on line B, 18 would be defective. This gives a total of 34 defective parts, with a 16/34 = 47% chance that the defective part in question came from line A.
3) Take event B as the event of drawing a heart on the first draw (P(B) = 13/52 = .25). Take event A | B as the event of drawing a heart on the second draw, with the condition that event B has happened (P(A | B) = 12/51 = .2353). By the general multiplication rule, P(A & B) = P(A | B) * P(B) = .25 * .2353 = .0588
4) What other methods are there to solving problem 2 besides the tree diagram method that I employed? It seemed the most intuitive to me, but also I felt like there was a faster method while I went through it.
1) P(B|A) would be greater, because every Engineering student has taken some type of calculus at some point in college. Therefore given the fact that the student is an Engineering student, the probability that they took calculus is 1 (or very close). Whereas given the fact that the student took calculus, the probability that the student is also an engineering student is not one, because not all students who take calculus are engineering students.
2) There is a 47.01% chance that the part is picked from Line A given that it is defective.
3) There is a 5.88% chance that both cards are hearts.
1.
P(B | A) is greater. It reads the probability of a student taking calculus given that he/she is an engineering major. This is bigger because all engineering students take calculus, but not all people who have taken calculus are engineering majors.
2.
P(Def) = P(A is Def.) + P(B is Def.) = 0.016 + 0.018 = 0.034
P(A | Def.) = P(A & Def.) / P(Def.) = (0.016) / (0.034) = 0.471
So there is a 47.1% probability it came from line A
3.
P(A | B) = P(A | B) * P(B) = 12/51 * 1/4 = 0.0588 = 5.88%
4.
I’m still sort of confused over the general multiplication rule, could you go over more examples?
1) Assuming all engineering majors take calculus in college since it is required, P (B | A) is higher. P(A | B) is lower because there are several people who take calculus in college but are not engineers.
2) P (Made on Line A | is defective) = P (Made on line A & defective) / P(is defective).
P(Made on line A & defective) = .4*.03 P(is defective) = .4*.03+ .6*.04
P = .012/.036
P = 1/3
3) 1/52 * 1/51 = .000377
4) How did they find the rule for compliments?
1. P(B|A) is greater, since there are more students that take Calculus than students that are engineering majors.
2. 53%
3. 5.9%
4. It pretty much makes sense.
1. P(A|B) is greater than P(B|A). If we looked at the number of people who are engineering majors and took calculus. Then divide that number by the number of people who took calculus, then we would get P(B|A). This number is lower than the number of engineering majors who took calculus divided by the number of engineering numbers. This is because the event of “taking calculus” is greater than the amount of engineering majors. The result can be concluded because every engineer must take calculus. So the result when dividing by engineering majors should be 1, while dividing by the number of people who took calculus would be less than 1, because people of other majors take the class.
2. The probability of a defective product from line A is computed by multiplying the probability that the product is from line A by the probability that the product is defective from line A. This probability is 0.4*0.04=0.016. The same can be done for line B, which comes out to be 0.6*0.03= 0.018. The probability of the product is defective is the result of adding these two probabilities which is 0.018+0.016= 0.034. The probability the defective product is from line A is 0.016/0.034 which is 0.4706.
3. The probability of both cards being hearts is calculated by multiplying the chance of picking a heart out of 52 by the chances of picking a heart from a deck of 51 cards with one missing heart card. It is (13/52)*(12/51)=0.058824
4. Is there a program that allows the user to insert a tree diagram along with its probabilities so that it is easier to visualize the data?
1) Suppose you select a random member of the Vanderbilt class of 2011. Let A be the event that the student was an engineering major, and let B be the event that the student took calculus while in college. Which is greater, P(A | B) or P(B | A)? Why?
> I would say P(B | A) because all engineering majors are required to take calculus in college (although some test out) but not all calc students are engineering majors. Other majors also take calculus.
2) Suppose a factory has two production lines, A and B. Line A produces 40% of the factory’s output, and Line B produces the rest. Of the parts made on Line A, 4% are defective. Of the parts made on Line B, 3% are defective. Suppose a part is picked at random from inventory and found to be defective. What’s the probability it came from line A?
> .47
3) Suppose you draw a card from a standard 52-card deck at random, then draw a second card at random from the remaining 51 cards. Use the General Multiplication Rule to find the probability that both cards are hearts.
P(A&B) = P(A|B) * P(B)
= (12/51) * (13/52)
= .059
4) What’s one question you have about the reading?
> The general multiplication rule.
1) P(B l A) because the chances that an engineering major has taken calculus is very high since almost every engineer is required to take it. P(A l B) would be less because it takes into account a lot of arts and science students.
2) 3.4 out of 100 parts from both machines will be defective and 1.6 of 40 parts coming from “A” will be defective. So 1.6/3.4 = .4705 so 47.1%
3) The probability that the second card will be a heart given that the first card is a heart is (1/4) x (12/51) = 1/17 and then using the multiplication rule (1/17) x (12/52)= (4/289) or 1.4%
4) Is there a limit as to how many rows and columns you can have in a probability table? I would think not but just curious.
1. Suppose you select a random member of the Vanderbilt class of 2011. Let A be the event that the student was an engineering major, and let B be the event that the student took calculus while in college. Which is greater, P(A | B) or P(B | A)? Why?
P(B|A) because if you know the student is an engineer, calculus is required it would be almost 100%. However, if you know the student took calculus, they could be a variety of different majors.
2. Suppose a factory has two production lines, A and B. Line A produces 40% of the factory’s output, and Line B produces the rest. Of the parts made on Line A, 4% are defective. Of the parts made on Line B, 3% are defective. Suppose a part is picked at random from inventory and found to be defective. What’s the probability it came from line A?
Assuming 100 parts are made, 40*.04= 1.6 defective from A, and 60*.03=1.8 defective from B. Since we know the part is defective (1.6/(1.8+1.6)) = 47% chance that the part is from A.
3.Suppose you draw a card from a standard 52-card deck at random, then draw a second card at random from the remaining 51 cards. Use the General Multiplication Rule to find the probability that both cards are hearts.
P(A and B) = (13/52)*(12/51) = 5.88%
4. What’s one question you have about the reading?
What kind of probabilities do you get when there is no correlation between the event and the condition?
1. P(B | A): all Engineering students have to take 155A/B and 175. Maybe 2/3 of the population(Really have no idea here) of ANS/Peabody/Blair Take calc.
2. P(A&D)=.016 P(B&D)=.018
P(A | D)=(.016/(.016+.018))=.471
3. (13/52)*(12/51)=.0588
4. How does one compare his/her poker hand to the possible hands of his/her opponents to see how good of a chance s/he has of winning?
1.) P(B|A) is greater than P(A|B) because P(B|A) is a probability that the student took calculus given the condition that the student is an engineering major, and almost all engineering students take calculus. On the other hand, P(A|B) is the probability that the student is an engineering major if the student took calculus. A lot of students with majors other than engineering take calculus; like maths, physics, chemistry, etc.
2.) 0.667
3.) 0.059
4.) none
1. Suppose you select a random member of the Vanderbilt class of 2011. Let A be the event that the student was an engineering major, and let B be the event that the student took calculus while in college. Which is greater, P(A | B) or P(B | A)? Why?
P(B|A) is greater because all engineers must take calculus whereas not all calculus students are engineers.
Suppose a factory has two production lines, A and B. Line A produces 40% of the factory’s output, and Line B produces the rest. Of the parts made on Line A, 4% are defective. Of the parts made on Line B, 3% are defective. Suppose a part is picked at random from inventory and found to be defective. What’s the probability it came from line A?
Say 1,000 parts were made. Of these 1,000, 400 of them would have been made by machine A, and 16 of these would have been defective. 600 parts would have been made by machine B and 18 would have been defective. Therefore, the likelihood that a defective part came from machine A is 16/34, or 47%.
Suppose you draw a card from a standard 52-card deck at random, then draw a second card at random from the remaining 51 cards. Use the General Multiplication Rule to find the probability that both cards are hearts.
The probability of the first card being a heart is 1/4. The probability of the second card being a heart is 1/4. The probability of both cards being hearts is 1/4*1/4, or 1/16.
What’s one question you have about the reading?
I didn’t have any questions.
1) Because event A includes event B (All engineers must take at least 14 hours of calculus) P(A) > P(B) thus P(A | B) > P(B | A)
2) Take a situation that remains consistent with the ratios, where A produces 400 units, and B produces 600 units. .04*400 = 16, while .03*600 = 18, thus in all cases it’s more probable that unit B is producing faulty units
3) probability of picking a heart once is .25. Picking another is 12/51 = .235. .25*.235 = .059 ~ a 6 % chance
4) Does P(A | A) mean P(A & B) / P(A) ?
1) P(A|B) is greater because all engineering students need to take calculus in college and it is not necessary for students who take calculus is an engineering major
2)The probability is 0.04
3)(13/52)*(12/51)=0.0588
4)The P(A|B) concept is still confuses me
1. P(B|A) is greater than P(A|B) because calculus is a required course to be in an engineering major, while other students taking calculus might be majoring in Math or other non-engineering majors.
2. P(defect) = .016 + .018 = .034
P(line A|defect) = .016/.034 = .4706
3. P(second heart & first heart) = .235 * .25 = .0588
4. I am a bit confused over marginal and joint probabilities.
1. P(A|B) because all engineers have to take calculus.
2. 47%
3. 5.88%
4. nope
1) P(B|A) is greater. The probability that a engineering student took calculus is greater than the probability that a student who took calculus is an engineer.
2) 0.898
3) 0.0588
4) How do the remarks on the sum of conditional probabilities apply to a variable or process with infinite values or outcomes?
1. P(B | A) is greater because all engineers are required to take calculus. Only a small few test out due to AP credit.
2. P(A) = (0.40)*(0.04) = 0.016
3. (13/52)*(12/51)=0.0588
4. I’m confused on the difference between joint and conditional probability. From what I understand:
Joint: the probability that X and Y occurred (1 step?)
Conditional: the probability that X occurred because Y occurred (2 steps?)
The example used to explain joint probability (drug use of parents and children) seemed to imply causation. In that example:
Joint: What is the probability that both parent and child used drugs?
Conditional: What is the probability that the child used drugs if the parent used drugs?
And because of causation these questions seem the same to me.
P(B|A) would be greater because the amount of students who are engineers are included in the number of students who take calculus. Therefore B will always be greater than A and results in a larger probability of the chosen student taking calculus.
I don’t understand how to do number 2
0.037% both cards are hearts
1. P(B|A) is greater, because more students take Calculus than major in engineering4.
2. 47%
3. 5.9%
1. P(A | B) = P(A&B)/P(B). Since all engineers take calculus, but not all people who take calculus are engineers, P(B)>P(A). Therefore, P(B | A) > P(A | B).
2. P(A | defective) = P(A&defective)/P(defective)=0.0016/0.0034=0.4706
3. (14/52)*(13/51)=0.0686
4. What is the difference between P(A & B) and P(A And B), because with the equation the book gives us for P(A And B), the P(B)s cancel out and you’re left with P(A & B)?
1. There are two probabilities. P(A | B) represents the probability that someone is an engineering major given that they took calculus while in college. P(B | A) represents the probability that someone took calculus in college given that they are an engineering major. Considering that all engineering majors are required to take calculus in college but not all calculus students are required to be engineering majors, it is clear that P(B | A) > P(A | B).
2. P (Line A | Defective) = P (Line A & Defective) / P (Defective) = .4706
3. P (A and B) = P (A | B) * P (B) = (13/52) * (12/51) = 0.0588
4. Why would we use anything but tree diagrams? They seem to be the truly superior method to solving probability problems.
1. P(B | A) because most engineering students take calculus.
2. 67%
3. 5.89%
4. The reading was very clear.
1. P(B|A) is greater. Because many non engineering students have to take calculus, when you calculate this number the smaller event (A) will be the divisor.
2. 4% of the 40% that is produced on Line A is defective, accounting for .4*.04 of all units, leading to 1.6% of units all together that are defective. 3% of 60% is .04*.6 = .018 or 1.8% of all parts are defective as a result from Line B. As 3.4% of all units are defective and 1.6% are from Line A, picking a random defective unit has a 42% chance of coming from Line A.
3. P (H1 and H2) = P(H1 | H2) * P(H2) = P (H1 & H2) / P(H2) * P(H2) = 1/4 * 12/51 = 5.88%
4. Not really a question, just a little fuzzy about this material. Would like to go over it in lecture more.
1. The probabilities are the same because these two are actually the same event.
2. Let the factory output be 1000. Then A produces 16 defectives and B produces 18 defectives. Therefore the probability that a defective part was produced by A would be 16/34 = 0.47
3.There are 13 hearts in the deck of 52, so the first pick will be 13/52 = 0.25
There are 12 hearts left in the deck of 51, so the second pick will be 12/51 = 0.235
Therefore, the overall probability would be 0.25 * 0.235 = 0.059
4. Does A & B has to be independent in order to calculate P(A | B)?
1. P(B|A); using the conditional probability formula P(A|B) and P(B|A) have the same numerator but the denominator of P(B|A) is smaller. (Assuming that the two events are independent)
2. P(came from A|defective) = P(A&def)/P(def) = .016/.034 = .47059; about 47.1%
3. P(A|B)*P(B) = (12/51)*(13/52)/(13/52) * (13/52) = 0.0588235294 = about 5.9%
4. Would we be able to answer (1) if the events in (1) were not independent (i.e. being an engineering major determines whether or not a student will take calculus)?
1. P(B | A) is greater because all Engineering students have to take Calculus, while only some of the people who take Calculus are Engineers
2. 47.06%
3. 1/17
4. What is a good substitute for a tree diagram if there are too many possibilities to be able to easily draw out all possible situations?
1. P(B|A) is greater since most, if not all, engineering students take calculus while not all students who took calculus are engineering students.
2.47.06%
3.1.775%
4.In the case of joint probabilities, is the a way other than Venn diagrams to express them graphically?
1. P(A | B) represents the probability that a student is an engineering major given that they took calculus. P ( B | A) is the probability that a student took calculus given that they’re an engineering major. Because 100% of engineering students take calculus, but not all students that take calculus are engineering majors, P (B | A) is greater (assuming no engineering students placed out of calculus from AP credit).
2. 47%
3. P( A & B = hearts) = P (B = heart | A = heart) * P (A = heart) = (12/51) * (1/4) = 5.88 %
4. if P (A & B) = P (A | B) * P (B) does P (A & B) = P (B | A) * P (A)?
In response to your question, Yes.
1. Using the bayes formula, the two fractions representing the two probabilities have the same numerator, therefore the one that has the smallest denominator Is the one that represents the smaller probability. And since the probability of students that take calculus is higher than the probability of students in engineering because all engineering student have to take calculus and some non engineering students take calculus too. Therefore the P(B|A) is greater because the fraction used to find it has a smaller denominator.
2. P(part defective| part came from A) = P( part is defective and part came from A)/P(part came from A) = 1%/40% = 2.5%
3. P = (13/52)*(12/51) = 5.88%
4. Don’t understand the bayes theorem entirely
Suppose you select a random member of the Vanderbilt class of 2011. Let A be the event that the student was an engineering major, and let B be the event that the student took calculus while in college. Which is greater, P(A | B) or P(B | A)? Why?
The latter, since given that the student is in engineering, it is very likely they took calculus.
Suppose a factory has two production lines, A and B. Line A produces 40% of the factory’s output, and Line B produces the rest. Of the parts made on Line A, 4% are defective. Of the parts made on Line B, 3% are defective. Suppose a part is picked at random from inventory and found to be defective. What’s the probability it came from line A?
prob = 1.6 / (1.6 + 1.8) = 8/17 = prob total defective from A / prob total defective
Suppose you draw a card from a standard 52-card deck at random, then draw a second card at random from the remaining 51 cards. Use the General Multiplication Rule to find the probability that both cards are hearts.
ans = 13/52 * 12/51
What’s one question you have about the reading?
Further explanation of using the tree diagram.
1.) P(A|B) = P(Engineering major | took calculus) and P(B|A) = P(took calculus | Engineering major). Notwithstanding AP credits or something, P(B|A) should be greater, because all engineering majors are required to take calculus, but not all students required to take calculus are engineering majors.
2.) From the tree diagram, we can calculate this probability as [ (0.4)(0.04) / ( (0.4)(0.04) + (0.4)(0.96) ) ] = 0.471. So there is a 47.1% chance that the defective part came from assembly line A.
3.) There are initially 13 hearts in the deck, so the probability of getting a heart on the first draw is 13/52. After this event, there are only 12 hearts left in a deck of 51 cards, so the probability of getting a heart on the second draw is 12/51. The probability of getting a heart both times is thus (13/52)(12/51) = 0.0588. So there is about a 5.9% chance of getting two hearts in a row on the first two draws.
4.) The sum of the conditional probabilities for events A_1 through A_k given B seems trivial. It seems to amount to saying nothing more than the probabilities of all the disjoint events A_1 through A_k must add up to 1. Is there anything more to the rule than this?
In response to your question, No.
1. P(A | B) is greater than P(B | A)…. It is a requirement that all engineers take calculus (although a select few may test out of all calculus due to IB or AP credit). In addition, more than just engineering students take calculus (Any major can take calculus, nearly all students). P(A | B) reads as, the probability of A given B. So if it is given that the student is an engineering student, then there is nearly a 100% chance that the student took calculus in college. While, P(B | A) reads as, the probability of B given A. Lets estimate that 70% of Vanderbilt students take calculus, given that the student took calculus in college, the prob of that student being an engineering student would be closer to 27.28%. Or the # of engineering students/number of Vanderbilt students that take calculus (I’m assuming about 70% must take it). (.2728=1302/(6817*.7))
2. The probability that a RANDOM part is from A (.4). Prob that a part from A is defective is (.04). So the probability that it is both from A & Defective is then (.4*.04)=.016, or 1.6%.
3. In a deck of 52 cards there are 4 different suits. 52/4=13 cards of each suit. The general multiplication rule states that P(A and B)= P(A|B)*P(B). The probability that the first card is hearts (P(B)) is 13/52=25%. Given that the first card is hearts, there are now only 12 hearts left in the deck of 51 cards. So the probability that the second card is hearts given that the first card is hearts is 12/51=23.53%. So the probability of having both cards be hearts is .25 * .2353= .0588 or 5.88%.
4. I didn’t understand the Sum of conditional probabilities.. Or rather I don’t know if i do because it seems too simple. Does it simply mean that the sum of the probabilities of all possibly outcomes is 1 (or 100%)?
1) P( B | A) is greater. This refers to the number of students who have taken calculus with the condition that they are an engineering major. All engineering majors take calculus. P(A | B) is less because just because many students take calculus. That doesn’t make them all engineering majors.
2) 0.47
3) 0.059
4) Number 2 of these questions
1) P(B | A) is greater since most engineering majors will take calculus, but not many calculus takers are engineering majors as many are science majors.
2) P(Made on A and Defective) = .4*.04 = .016
P(Made on B and Defective) = .6*.03 = .018
P(Made on A | Defective) = .016/(.016 + .018) = .471
3) 13*12/(52*51) = .059
4) How do we incorporate error into probability calculations?
1. P(A|B) is asking for the probability a student is an engineering major, given that they have taken calculus. P(B|A) is asking for the probability that a student has taken calculus, given that he is an engineering major. Since virtually all engineering majors are required to take calculus, it seems that P(B|A) would be greater.
2.For explanation purposes, assume 100 units are produced. 40 produced by A, 60 by B. This means that from line A, 4%, or 1.6 units will be defective. For B, 3%, or 1.8 units will be defective. In order to determine the probability the selected unit came from A, compute 1.6/(1.6+1.8) = .47 or 47% chance.
3. Splitting these up into separate events, we know that event A has a probability of 13/52 or 25%. The second event B has a probability of 12/51 or 23.53% chance. Multiplying these two together, we get .0588 or 5.88%.
4. The reading was pretty straight forward.
1. P(B|A) is greater, since more students that take calc than are engineering majors
2. 47%
3. 5.9%
4. Are all probability calculations error-free?
1.P(B|A) would be greater. Calculus is a requirement for all engineering majors. So, the probability that the student took calc given that the student is an engineering major would be 1. However, not every student who takes calculus is an engineer. So, the probability that a student is an engineering major given that the student is in a calc class would be less than 1 but greater than 0.
2. d = defective, ~d = not defective
P(A) = 0.4,
P(A&d) = 0.4*0.04 = 0.016,
P(d) = P(A&d)+P(B&d) = 0.016 +0.6*0.03 = 0.034,
P(A|d) = P(A&d)/P(d) = 0.4706
3.P(h1) = 1/4 = .25
P(h2|h1) = 12/51 = .2353
P(h1 & h2) = P(h2|h1)*P(h1) = .0588
4.What’s one question you have about the reading?
none
1.
P(B|A) is greater than P(A|B).
P(B|A) describes the probability that a given student took calculus in high school Given That they are an engineering major. Considering that calculus in high school was borderline required for engineering students, this is likely a very high figure.
The opposite (probability of being an engineer given they took calculus in high school) is likely much lower, considering many science majors would also have taken calculus in high school, and that the general caliber of students at Vanderbilt could suggest many humanities students may also have taken it.
2.
100% 40 % A – 4 % Defective = .40 * .04 = 0.016
– 96 % Fine =
60 % B ~ 3 % Defective = .60 * .03 = 0.018
~ 97 % Fine =
P(A|Defective) = 0.016 / (0.016+0.018) = .47059 = 47.1 %
3.
P(Hearts 1 & Hearts 2) = P(Hearts 2 | Hearts 1) * P(Hearts 1) =
P(Hearts 2 | Hearts 1) = 12 / 51 = .2352
P(Hearts 1) = 13 / 52 = .25
= .2352 * .25 = .05882 = 5.9 %
4. What’s one question you have about the reading?
Is it possible to restructure a tree to effectively display both (A given B) and (B given A), or should two trees be constructed, or is there a different display technique that is overall better at that?
1. P(B | A) is larger. In reality, almost all the engineering students are required to take the calculus course. Therefore this probability tends to be 1.
2.16/(16+18) = 47%
3. (13*12)/(51*52) = 5.9%
4. ..How can I find the answer of the above questions. I am not sure whether I am right or not…
P(B|A) -almost all engineers take calculus but not all students who take calculus are engineers.
.4 x .04 = 1.6%
(13/52)*(12/51) = 5.9%
I think I understood this section pretty well.
1. P(B|A) would be greater because all engineering students are required to take calculus of many different levels, whereas not all students that take calculus are engineering majors.
2. The probability would be 47.1% that the part came from Line A. ((.04*.4)/(.04*.4+.03*.6)=.471)
3. The probability would be 5.9%. (P(A and B)=P(A|B)*P(B) => P(A and B)=(12/51)*(13/52) =>.0588 ~> 5.9%)
4. In example 2.54, where did the number 85.88% come from? Was it purely random? Or was there a calculation that I’m not getting that it was derived from?
1. P(B|A) is greater than P(A|B). In both cases, the numerator is the same (A&B = B&A). However, it is expected that the number of students that take calculus exceeds the number of students that take engineering. Thus the denominator in P(A|B) is greater than that in P(B|A), making P(B|A) the higher probability.
2. The chance that it came from line A is 4% of 40 = 1.6% or 0.016.
3. P(A&B) = 12/13 * 13/52 = 12/52.
4. I need a better explanation of the general multiplication rule.
1. P(B|A) — since it is calculated as P(A&B)/P(B), P(B|A) is smaller than P(A|B) because it is more likely that a student has taken calculus than that the student is an engineering major, so a greater denominator would make the P(A|B) value smaller.
2. 29.4%
3. 13/204
4. what do you do in the case where conditions are not independent? for example — in the card example (#3), how do you still calculate probability when you do not know what first card drawn was?
A)
As far as I know, P(B|A) is greater than P(A|B) because the number of engineering students is less than the number of students. who took Calculus (Because the class is required for engineerning and non engineering students)
B)
This is little trickey.
1) 4% of 40 = 1.6
2) 3% of 60 = 1.8
3) From that, it is clear that the chance of defective item from line A is smaller than the chance of defective item from line B.
4) p = (1.6/(1.6+1.8)) ==> = 47% chance of defective item from line A
C)
1) We have 13 heart cards under normal situation. Also, deck card has 52 cards under normal situation.
2) after withdraw a card (assuming it is a heart card), we have 12 heard cards and a total of 51 cards.
3) Now, [(13*12)/(52*51)] ===> 5.88%
4)
The material is getting little bit confusing. I have some issue with imagining the sets of probability. I am confused with different between P(A|B) and P(B|A).
1. P(B|A) is greater because while it is likely that the student could have taken calculus in college, it is required of all engineering students, so it would make the probability 100%.
2. P(came from A | defective) = P(came from A & defective) / P(defective) = 0.016/0.034 = 47.06%
3. P(first card is hearts & second card is hearts) = P(second card is hearts | first card is hearts) * P(first card is hearts) = (12/51) * (13/52) = 0.0588 or 5.88%.
4. I don’t really understand why the General Multiplication Rule works theoretically.
1. P(B | A) because there are going to be more students that take calculus that aren’t just engineering majors.
2. 2) P(A and Defective) = .4*.04 = .016
P(B and Defective) = .6*.03 = .018
P( A | Defective) = .016/(.016 + .018) * 100 = 47%
3. 13*12/(52*51) *100= 59%
4. I am having trouble grasping the concept behind conditional probability.
1. It is known that all engineers must take calculus at some point in college. Therefore, the probability that a student took calculus given that the student is an engineering major is greater than the probability that the student is an engineering major given that the student took calculus.
2. The probability it came from A is 16% (0.40*0.04 = 0.16), versus the probability it came from B, which is 1.8%
3. (1/13)*(1/12) = 0.0064 = 0.64% probability that both cards are hearts.
4. Can we use some sort of probability density visualization in order to better grasp the concept?