difficult child brought home homework. Amazing.... Actually asked for help...gasp.... (The world must be coming to an end. ) Math never was my strong suit. No textbook sent home, so no way to look up info to set up formula.... Anyone know how to set the formula up to "Find the probability of each event?" 1) rolling a 1, 2, or 3 on the first roll of a 1-6 number cube and rolling a 4, 5, or 6 on the second roll of the same cube. 2) Randy has 4 pennies, 2 nickels and 3 dimes in his pocket. If he handomly chooses 2 coins, what is the probability that both are pennies? I think I've got it, but if they are wrong, I'll never hear the end of it...lol.

These are the answers my "math whiz" kids got: 1) one-half probability that 1, 2 or 3 will be rolled on the first try, and one-half probability that 4, 5 or 6 will be rolled on the second try (3 numbers out of 6 numbers each time) 2) four-ninths probability that a penny will be chosen the first time, and three-eighths probability that a penny will be chosen the second time. My son said to multiply those numbers together to get a 16.7 percent probability of choosing two pennies Hope that's right!

Hey Sheila, I'm good for something! smallworld's kids are exactly right. Here is how to think it through: In the first problem, the events are independent of each other. In other words, the probability of the second event occuring is not altered by the first event. Since you want to roll a 1,2,3 on the first roll (3 favorable outcomes out of 6 possible outcomes) which is 3/6 or 1/2 "and" you want to roll a 4,5,6 on the second roll (3 favorable outcomes out of 6) which is 3/6 or 1/2, you multiply the probabilities together and 1/2 times 1/2 is 1/4 or a 25% probability. Now, in the second problem, once you pick a penny the same penny cannot be picked again to the second probiblity is dependent on the first. So you have to approach it a little differently. The first probability is 4 pennies out of 9 coins which is a 4/9 probability. The second probability is 3 remaining pennies out of 8 remaining coins or 3/8. Again, you multiply the probabilities together so 4/9 times 3/8 is a 1/6 chance or a 16.7% probability. Way to go, smallworld's math whiz kids!! ~Kathy

Ha!! I'm a math teacher, but probability is my worst area. But, Kathy and whiz kid have it right. On a side note, my best friend's daughter is in 6th grade, which is what I teach. She has these problems of the week. My gosh...are they trying to kill the kid??? My husband, who has a PhD in physics had to take the darn thing home to figure it out. Supposedly the purpose is not the answer, but the process, which they have to write down. To me, it was a rather humiliating exercise and I wouldn't have received any benefit as a child in doing it. They have one problem every week. Geez... Abbey

I loved probability theory both at school and at uni, where the theory was applied to studying populations of tagged/untagged specimens in the wild. The answers are correct - we often get bamboozled by probability, thinking that everything in the world is connected far more than it is. We might sit at a poker machine (slot machine or whatever you call it) and pour in coin after coin and think, "I've got to get a payoff soon, it's got to be my turn soon." But in reality, you have no better chance with each coin than with all the ones before it. As Kathy said, this is random. We just like to hope that it's not. The classic, simple question is: "A kid tosses a coin and gets 99 heads in a row. What is the chance that the next toss will also be a head?" The answer is, 1 in 2. Or, 0.5. However, in practical terms, I'd be checking the coin to make sure there is no bias, such as having two heads. If the coin DOES have two heads, then the next coin toss is a certainty to get a head. And yes, to get the probability of consecutive events, you multiply them together, but sometimes you have to be slightly sneaky about it and multiply the probabilities of the events NOT happening, then subtract your final answer from 1. The second example is an interesting one - it's "sampling without replacement". If you follow it through, the more you sample without replacement, the better your chances each time of getting it right. The safe in "The Price Is Right" is a good example - it has three knobs. Each knob has three numbers. You can't use the same number twice (which makes it sampling without replacement). The chance of getting the first knob right is 1 in 3. The chance of getting the second knob right is 1 in 2. The last knob has no choices left, but the remaining number, so if the first two are right, then this last number is automatically correct - probability 1. The total probability is one third times one half, or 1 in 6. The more numbers you have to choose from (and the more knobs), the process simply continues. You can see that for 10 knobs and 10 digits, you're multiplying 1 over 10 x 9 x 8 x 7 x 6 x 5..etc. This is more easily written as 10! or 10 factorial. Writing it as 10! (or whatever number) saves paper. So if you ever see that little ! after a number, that's what it means. It is NOT mathematicians becoming excited. We use "sampling WITH replacement" whenever tagged animals are released into the wild. Have you ever wondered why scientists bother to tag animals? You can't tag 'em all, and how can you get useful information from just half a dozen tagged crocodiles, for example? This was back in the days when animals were tagged without radio tracking - the tag had to be attached in a way that made recapture as likely as capture of a fresh, previously uncaptured specimen. Let's say we capture and tag 20 specimens, then release them. We lay nets again, and capture another 20. Of that 20, 5 wear tags (and are therefore recaptures). What is the total population in the wild? It's actually not difficult, if you think about it logically. The group captured is (hopefully) a fully random representative sample. Of the ones we've just captured, 5 wear tags. This means we tagged 5 out of 20, or 20%, of the population. Therefore 20 specimens tagged first go, is 20% of the population. Total population - about 100. You can't be exact because there are always some random factors. Is the tag making it easier to recapture them? Are the ones originally captured more stupid than most (and hence blunder into the nets again)? I remember one case where two blue wrens were recaptured, seven years later, in exactly the same place where they had been first captured and tagged, as yearlings. They simply hadn't changed territory or behaviour patterns. This sort of thing means that where mathematics meets zoology, life can become complicated and unmathematical. So sampling with replacement, and sampling without replacement - it all begins with coins in the pocket. And here's a cute brain teaser for small difficult children - you're in your bedroom and it's dark. You need to get a pair of socks from your sock drawer, but you can't turn on the light. You don't want to grab ALL the socks, you want to get as few as possible. In the drawer are loose grey and brown socks, all jumbled together. You don't care whether you get a pair of grey, or a pair of brown, just so long as you have a pair. All the socks are identical, other than these two colours. What is the minimum number of socks you need to get, to be certain you have a pair? Marg

Bless you all! I got the 1st one right; but blew the second one. I was thinking one would get two coins at a time. lol

One more. I've confused myself with "dependent" and "independent"... There are 20 true/false questions on a test. You do not kow the answer to 4 of the questions, so you guess. What is the probability that you will get all 4 answers right? I have doubts difficult child will ever "get" this kind of reasoning. Example: Chance of selecting 2 pennies from his pocket is 100% because he can feel the difference in the coins. So I tell him to pretend he can't feel the coins. You got it, "but I can, so the probability is 100%." What do you tell the student to get them past this concrete thinking???????????

Sheila, the question above is independent. Question 1's answer is not dependent on question 2's, question 3's or question 4's answers (in other words, there'a 50 percent chance on each question that it's true or false independent of the other questions' answers). So the answer is 0.5 X 0.5 X 0.5 X 0.5 = 0.0625 or 6.25 percent probability of getting all four questions right.

Sheila ~ this is independent probabilities. Guessing one answer right or wrong does not have any effect on the next guess. So there is one favorable outcome out of two possible outcomes on a true/false test, or a 1/2 probability that you guess each question correctly. Since you are guessing on 4 questions you would multiply 1/2 times 1/2 times 1/2 times 1/2 and get a 1/16 probability (6.25%) of guessing the four answers correctly. The fact that there are 20 questions is just extraneous information. As far as abstract thinking, I believe that everyone reaches that at a different point in their lives. In fact, that is the reason that I have a problem with moving Algebra 1 down to younger and younger ages. I think many students are not ready for the abstract thinking needed for higher math and we are setting them up for failure. by the way, lots of people have problems with probability. I hate to teach that chapter since many students have a hard time with it. I hope I helped. ~Kathy

Some things never change. I worked the problem correctly the first time. Then "overanalyzed" by considering the answer based on dependent events. I use to tell myself, "You first answer is usually the right one." Kathy, you have been helpful. Besides the math support, I always try to gage how things are going with difficult child based on how his age-group/grade handles similar situations or environments. When it pertains to education, I assume that if his peers can master the content, difficult child should be able to do so as well within a certain range (fair to good). It's some relief to know that other students don't automatically get this type reasoning. But with difficult child, you can't get him past black/white thinking. Thank goodness there's more aspects of math than probability or most likely, a lot of us would be in serious trouble -- me included. lol Thanks again! by the way, difficult child's struggling with introductory algebra. Doesn't bode well for the future I fear...

Well, Sheila, if it makes you feel any better, I struggled with 8th grade math. The lightbulb didn't go off until Algebra 2 when it all started to make sense to me. And look where I ended up! I'm sure my ninth grade Algebra 1 teacher would be surprised. ~Kathy

Just reporting in that with everyone's help, difficult child's grade in Math rose from an "F" to a "D." Thank you!!! Kathy -- that gives me hope. I sure don't expect him to end up a math teacher, but maybe we'll make it through high school. lol Strange how my expectations have changed. First I made plans for college, then the goal was to get through high school. Today, I'm having trouble with real expectations of completing this year (7th grade) with any type success. It's way past time for school to be out for the year. difficult child, his teachers, and me -- we NEED for school to be out. lol Don't know what they pay teachers who teach kids in puberty, but I know it can't possibly be enough! There should be some type of hazard pay involved! Thanks again, ladies! You all did good on this assignment!

Sheila, Math is a tricky devil. I was HORRIBLE at it from about grade 6 on. It wasn't until college that it 'clicked.' I have been teaching math now for 21 years. Keep in mind that your brain is like any other organ in your body. Everyone developes at a different rate. He's at that age when you switch from concrete to abstract thought patterns. For some, it happens early...others (like me) it took awhile. It doesn't mean you're 'dumb,' it just means you haven't reached that point yet. It's like asking a newborn to walk when they haven't developed the skills to do so yet. (I'm hopping off my teacher podium right now.) Abbey

Sheila, I'm glad to hear that difficult child is passing math now and that I could help. Feel free to PM me if you ever have any other math questions. by the way, I mentioned in another thread that I am starting my postgraduate degree this summer. The first class that I am going to take is Teachers and the Law. I'm sure that I'll be popping in the Special Education forum for help. I hope that you won't mind my tapping into your expertise in that area. ~Kathy

Thanks for the offer! My knowledge of the law is limited to students with disabilities, but I'd be glad to help. I have a lot of resource material also -- might help save a minute or two of research time for you. I've taken various business and real estate law courses. Beware of the caveats. lol

Absolutely nothing to add, as math never was my strong suit (and trust me, THAT is the understatement of the year!). Still, it's been fun to read along, and to see how everything worked out. I have a mental block where math is concerned. It's like my brain freezes at the mention of the M word. Let alone the A (algebra) or even worse, the G or the T words. Barbara

<div class="ubbcode-block"><div class="ubbcode-header">Quote:</div><div class="ubbcode-body">Let alone the A (algebra) or even worse, the G or the T words. </div></div> You realize that you are talking about my life. ~Kathy

<div class="ubbcode-block"><div class="ubbcode-header">Quote:</div><div class="ubbcode-body">Marguerite wrote, And here's a cute brain teaser for small difficult children - you're in your bedroom and it's dark. You need to get a pair of socks from your sock drawer, but you can't turn on the light. You don't want to grab ALL the socks, you want to get as few as possible. In the drawer are loose grey and brown socks, all jumbled together. You don't care whether you get a pair of grey, or a pair of brown, just so long as you have a pair. All the socks are identical, other than these two colours. What is the minimum number of socks you need to get, to be certain you have a pair? </div></div> Marg, How'd you know what my sock drawer looks like? LOL A one time co-worker of mine solved the sock problem by only ever buying a particular brand and color of socks (black). That way any two socks at random were a match. Now, as to the problem: my intuitive answer right off the bat was "half + 1". But actually the answer is 3. If you get two they might match, or they might not. If not, the third would have to match one or the other. (I'm ashamed to say how long it took me to reason that out. I must be a small difficult child> In the real world, though, we have to deal with the problem of un-mated socks. You'd think that the mates to any odd socks would show up eventually, but it ain't so. Where do those odd socks' mates go? Last time we moved, I had about half a dozen un-mated socks. We cleaned out every drawer and closet in the house. The washer and dryer were empty and disconnected, no odd socks within or left in the utility room. We never threw away odd socks, on the assumption that they had to have a mate somewhere. So, where did they go? I think some kind of supernatural force is at work. That, or they were lost in laundromats and relative's houses on vacation. by the way, are grey and gray the same colo(u)r? I'm trying to learn Ozzie.