Solving probability problems

The reason that probability can encompass other categories so easily is that the probability of an event occurring is, at heart, a simple ratio: the number of desired outcomes/the number of total possible outcomes. To simplify matters, it can be helpful to break this ratio into its component parts. First find the total possible number of outcomes. Then find the number of desired outcomes. When we think about the issue this way, it seems much more manageable. Take this newer official question, for example: If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?  A) 1/4  B) 3/8 C) 1/2 D) 5/8 E) 3/4  On the surface, this is a probability question, but because we’re talking about divisibility, it’s also testing our knowledge of number properties. So let’s start by thinking about our total possible outcomes. There are 96 numbers between 1 and 96 inclusive, so clearly, there are 96 total possible outcomes when we select a number at random. We have the denominator of our fraction. Now we just have to figure out how many ways we can multiply three consecutive numbers, n(n+1)(n+2), to get a multiple of 8. Put another way, any multiple of 8, or 2^3, must contain three 2’s. One way this can happen is if the middle number, n+1, is odd, because every odd number must be sandwiched between a multiple of 2 and a multiple of 4. If n+1 is 3, for example, you’d have 2*3*4, which is a multiple of 8. (We need three 2’s in all. The 2 gives us one, and the 4 donates the other 2’s.) If n+1 is 5, you’d have 4*5*6, which is also a multiple of 8. (The 4 donates two 2’s and the 6 donates one. So long as we have three 2’s, we have a multiple of 8.) Between 1 and 96, we’ve got 48 odd numbers. The other way we can get a multiple of 8, when we multiply n(n+1)(n+2)  is if n + 1 is itself a multiple of 8. Clearly 7*8*9 will be a multiple of 8. As will 15*16*17. We can either count the multiples of 8 between 1 and 96, or we can use the trusty formula: [(High-Low)/Interval] + 1. The first multiple of 8 between 1 and 96 is 8. The largest is 96. And the interval will be 8. So we get [(96-8)/8] + 1 = 11 + 1 = 12 multiples of 8. So we have two categories of desired outcomes: there are 48 ways that n+1 can be odd, and there are 12 ways that n+1 can be a multiple of 8, giving us a total of 48 + 12 = 60 desired outcomes. We’re done! The number of desired outcomes/number of total possible outcomes is 60/96, which will reduce to 5/8. The correct answer is D. Takeaway: There’s no reason to be intimidated by probability questions, particularly when we remember that a probability calculation can be viewed as a ratio of two numbers. If we break the problem into its constituent parts, the question is often revealed to be quite a bit easier than it seems at first glance, a realization that proves true for almost any challenging  problem.