Another from my archives: Nice for Alg II level or so, I think.
A colleague from Colorado sent me an interesting probability problem the other day. I like it because it illustrates one of those serendipitous qualities of mathematics. Here is the problem. A Jar has a mixture of Red and White balls so that if you withdraw two, the probability of getting two alike, or two of different color are both equal to one-half. You may want to stop and try it before you read on.
Ok, so we let r = the number of red ones and w be the number of white ones, and the total is r+w. So how could we draw one of each color? Well, red first, and then white, or white first and then red. If we find the probability of each of these conditional events and add them up, that will have to equal 1/2. Ok, the …
Another from my archives: Nice for Alg II level or so, I think.
A colleague from Colorado sent me an interesting probability problem the other day. I like it because it illustrates one of those serendipitous qualities of mathematics. Here is the problem. A Jar has a mixture of Red and White balls so that if you withdraw two, the probability of getting two alike, or two of different color are both equal to one-half. You may want to stop and try it before you read on.
Ok, so we let r = the number of red ones and w be the number of white ones, and the total is r+w. So how could we draw one of each color? Well, red first, and then white, or white first and then red. If we find the probability of each of these conditional events and add them up, that will have to equal 1/2. Ok, the probability of red on the first draw is r/(r+w), and on the second ball the probability is w/(r+w-1) since one of the red balls will be missing. The opposite order is exactly the same with the w and r reversed, so the probability of getting one of each color is 2rw/[(r+w)(r+w-1)]. Setting that equal to 1/2 we get
If we expand (r+w)2 and subtract the 4rw we get 0=r2-2rw +w2-r-w. NOTICE the symmetry, we could exchange r and w and get the same equation. We know right off that any solution (r,w) will have another solution (w,r).
One of the things that is often hard for students is to think of one variable as a constant and the other as a variable. I like to use the word "pronumeral", like a pronoun only instead of him or her we say "that number". It is like a variable that doesn’t vary, we just don’t know what it is in a particular case. So think of w as if it were fixed. We have that many white balls in the jar and we are wondering how many red can be put in to make the problem work... see it.. w is a "fixed" unknown, but r is going to "vary". That makes the equation a quadratic in r; Ar2 +Br+c=0 where A=1, B= -2w-1, and C=w2-w. We can solve this using the quadratic formula, but if this solution is going to be a rational number, and the number of balls in a jar must be rational, then the discriminant, the expression under the squre root radical in the quadratic formula, B2-4AC, must be a perfect square. B2= 4w2+4w+1 and 4AC= 4w2-4w; so B2-4AC= 8W+1. If there is a rational solution, it must be when 8W+1 is a perfect square.
Wait, I know this one! That’s a problem from number theory. The numbers that make 8W+1 a perfect square are called triangular numbers; 1, 3, 6, 10, 15. They are the sum of the first n counting numbers. But a neat thing happens if we plug 1 in for W, the solution for r is 3.... and if we use 3 for w, r=6. Each time we substitute one of the triangular numbers into the quadratic, the next comes out as a solution. So the probability of drawing two balls of the same color, (or of two that are not alike) will equal 1/2 whenever the number of balls of each color are consecutive triangular numbers. A very geometric solution to a very algebraic question.