On any given winter's morning, the probability of your car starting is 65% and the probability of your parents car starting is 85%. Calculate the probability of each of the following events: a) Neither car starts b) Both cars start c) Both cars start or neither car starts d) Only one car starts One more... 14 beads are on a leather rope necklace. 4 beads are identical spheres, six others are identical cubes and the rest are identical cylinders. a) how many different necklaces can you create with the beads b) how many of these necklaces will be symmetrical through the centre Anyone up for the challenge?

Lets name some events. A-your car starting B-your parents' car starting A'-your car not starting B'-your parents' car not starting. Now, we can assign some probabilities to those events. P(A)=0.65 P(B)=0.85 P(A')=0.35 P(B')=0.15 Assuming the cars starting are independent events (which isn't stated in the problem, but conditional probability doesn't seem to make much sense here, so we'll go with independence).... a) Probability of neither car starting: We could say this as the probability of your car not starting AND your parents' car not starting. The use of the word and is our clue to use the multiplication rule, so: P(A' and B')= P(A')*P(B') P(A' and B')=(0.35)0.15) P(A' and B')= 0.0525 b) Probability of both cars starting: The probability of your car AND your parents' car starting. P(A and B)= (0.65)(.85) P(A and B)= 0.5525 c) Both cars start or neither car starts: Here, we need the addition rule, since we've use the word or. In words, the addition rule states that the probability of event A or event B happening is the sum of the two events minus the probability of them both happening (the sum of the events minus the intersection. In this case, there is no intersection between the two events joined by the word "or", (since it's impossible for both cars to start and both cars to not start at the same time), so we don't have to worry about the intersection. Also, since we've already figured out the probabilities of both cars starting and both cars not starting, we'll just steal the numbers from our earlier work. So, P[(A' and B') or (A and B)]= .0525+0.5525=0.605 d) Only one car starts: We can reword this to say the probability that your car starts or your parents' car starts. Once again, we need the addition rule. Remember from above, that the addition rule states that we need the sum of the probabilities, minus the probability of them both happening at the same time. Again, we already figured out that intersection in part b, so we won't reinvent the wheel. So, we have: P(A) + P(B) - P(A and B) = 0.65+0.85-0.5525=0.9475 I'll come back for the beads one...it's bed time for the kids.

Dear MM, 1. I wish you had been my math teacher in high school. If you explain things to your students orally as well as you did in this post, they are very lucky. 2. My brain just exploded a little.

Only one car starts... I read that as... your car starts and your parents doesn't or your car doesn't start and your parents does =P((A & B') or (A' & B))

Okay, there's two answers to these questions. Which answer you go with depends on how you define the word "different". We have multiples of the same beads, but are we considering each bead as "different", or do we consider all the beads of the same description as "same". The more technical way to put this is how many distinguishable outcomes are there? If every bead is different, even if it looks the same as others, we just have a simple counting problem using the multiplication principle (14*13*12*11*...*1). I think though, that we're going for distinguishable permutations, so we won't worry about the simple version. Lets define a few things. In the problem above, we have n=14 (n, with no subscript, is the TOTAL number of beads). Since I don't have mathtype, and I don't know how to use LaTeX, we'll pretend that the number after the next few n's are really subscripts. n1=4 (spheres), n2=6 (cubes) and n3=4 (cylinders). The formula for this type of problem is: n!/(n1*n2*n3) This could be expanded for any number of subgroups. So, we have: 14!/(4!*6!*4!) = 210210 possible unique combinations. Now, we want to know how many of them would be symmetrical down the center and I'm going to assume we want distinguishable outcomes. Again, we have 14 beads, with an even number of each type of bead. We only need to look at half of the necklace, since the other half is going to be the mirror image, so lets cut all the numbers in half. n=7, n1=2, n2=3, n3=2 So, we have: 7!/(2!*3!*2!) = 210

I don't know if I'd call myself a real mathematician. HMM...I had a brain fart on the symmetry part of it. Please tell me I'm right.....

For problem 2) As stated, this problem is much harder than what is being worked out here. To really do this problem one would need to use Burnside's Lemma. If you don't want to take rotational/reflection symmetries into account (which you should to get a correct answer), then mmswm's answer is correct.

For an example of what I'm talking about see page 490. Contemporary Abstract Algebra By Joseph A. Gallian

I assumed that for a HS class, we were just talking about simple, mirrored symmetry . I suppose I should have mentioned the assumptions I made.

That is probably the case since this is supposed to be for HS, but it should be stated in the problem that symmetry is not being considered. My students found this topic (Burnside's Lemma) fairly difficult the last time I taught Abstract Algebra.

Easy for me, but would blow my students out of the water. Question 1 a) 6.75% b) 55.25% c) 62% d) 39.5% Question 2 a) 96 b) 48 Now for my turn. Are you up for the challenge of a REAL world question? You are sampling monitoring wells around a landfill for lead contamination. Well no. 1 is the upgradient well (the ground water BEFORE it touches the waste.) Wells 2, 3, and 4 are downgradient wells (after the water has gone through the waste.) The data is shown for 4 sampling events. Which (if any) downgradient wells show evidence of contamination from the landfill? Well 1: 6ppm 12ppm 8ppm 9ppm Well 2: 10ppm 12ppm 7ppm 8ppm Well 3: 10ppm 11ppm 8ppm 10ppm Well 4: 8ppm 6ppm 12ppm 9ppm

Interesting...I'd love to know how you determined that the answers to #1)(a) was 6.75% and #2) was 96 and 48

I don't teach HS. And giving correct solutions to problems does help your students. If you want to give incorrect answers to your students then go ahead. I'll stick to doing it right.

Goodness gracious... There really is no need to get nasty here... Clearly, HMM is a math professor and has a very deep understanding of the subject. Arguably, he understands math better than most of the math teachers on this board. Content knowledge is a good thing, and knowing how to do detailed math is a rare gift that is best shared for public good. One of the travesties of public school instruction these days is how badly math is taught. Part of this is plain ignorance - you can't teach math if you don't know it well. Good content knowledge is vital for a math teacher. At the same time, this is a board for teachers, and we are K-12 educators for the most part. We need to understand that our knowledge is only as good as our ability to teach kids. While rigorous theorems are essential for academic math, the most important thing is how we can apply this great content knowledge in the classroom. To quote one of my old administrators: "It doesn't matter what you know. What matters is what your kids know after they leave you." To quote the mom of one of my students: "You know your stuff but you're like a professor in college. My kid's only 12."

You didn't clarify whether this was a poor teacher's car or a normally maintained one... And you're all wrong- The probability of both cars not starting increases exponentially on Mondays when you're already late for an extremely important appointment (likely after your beaded necklace broke and you tried to pick up all the spheres, cubes, and cylinders....)