Categories Probability Statistics & Probability Probability Problem (from SBAC) Post author By mpershan Post date June 4, 2013 9 Comments on Probability Problem (from SBAC) What say you all about this work? Thanks to Andrew for the submission! Share this:EmailPrint ← “You ought to get that probability at least once.” → Coin Flipping 9 replies on “Probability Problem (from SBAC)” I’m going to say that someone miscalculated the probability of 2 even numbers, or forgot about it when choosing the spinners. That last sentence is poorly worded as well. Those sure aren’t standard dice. I wonder how many of their sides are even? Also all the rigamarole about dragging seems unnecessary when we only want a response that’s a single number. Seems like this would work just fine as a multiple-choice question the way it is phrased. Help me out guys. I am getting 25% for the two even numbers and 40% if two regions are colored red. Doesn’t that result in a 10% probability of success? Ignore my previous post. I thought we were looking at the design of the question and not the posted solution. It has been a long day. My problem with all of the SBAC stuff is the over-reliance on nifty graphics and flash motion with dragging. As mentioned above, it’s unnecessary and often counter-productive. But someone is convinced that “kids today” need this to “stay engaged.” Just to throw another bit on the fire, the head of SBAC told us that they wouldn’t listen to criticism of the questions or the formatting because “These questions have been out there for months. They’ve been vetted by experts.” I say (1/5)(1/2)(1/2)=1/20 or 5%. My guess would be the mistake was using a straight 1/2 for the probability that both dice would be even instead of thinking of the fact that the probability of each dice being even is independent of the other so you have a 25% chance of two even numbers like Blaise stated. Where is the problem? In this exercise all there is the notion of independent events, and the part-whole relationship. Finally the most elemental numeracy solves the numeric relationship. 1) Probability (even face) = 1/2. Probability (red section) = n/5. Thus Probability (even & even & red) = 1/2 x 1/2 x n/5. And 1/2 x 1/2 x n/5 = 1/10 implies n = 2. 2) Following Laplace: – the number of all cases possible: 6 x 6 x 5, – the number of cases favorable: 3 x 3 x n, – and “10% of the players to win” means 1/10 = (3 x 3 x n)/(6 x 6 x 5), that makes n = 2. @fdejuan it is not the problem, per se, that is noteworthy. IMO, it is the expectation that a 8th grader would be able to do this problem without a steep background in probability. It is not realistic. I agree with Rebecca, that the student multiplied .2 and .5 by mistake. In that case it’s a fairly minor problem that’s easy to rectify. The other possibility of course is that the student has no clue how to handle either probability or proportion, and was just randomly assigning colours to the spinner. Hopefully that’s not the case, but there’s no way to tell. The issue to me is with the question format, as there’s no real way to discern student thinking. Call me a dinosaur, but this question should have been on paper, with a separate section for rough work. Comments are closed.