Vandal Science News-June 2016
A Newsletter for Alumni and Friends June 2016
While we are still in shock over the loss of Dean Paul Joyce, we’re also pressing forward in the excellent tradition that Dean Joyce helped to establish. Our faculty continue to excel and distinguish themselves – you can read about a few examples of that in this issue of the Vandal Science News.
Also, we just completed the university’s 121st commencement in May. That gave us an opportunity to honor the achievements of many of our outstanding graduating students. This newsletter always gives us an opportunity to show off some of the great things our students are doing.
The quiet of campus in the summer is really an illusion. There is great work being done during these months, and the excitement of doing science continues in our labs and offices as well as out in the field. We’ll reconvene in late August ready to greet a new class of Vandals and begin another academic year. We appreciate your continuing support of the College of Science.
Associate Dean Mark J. Nielsen
An Eye on Her Future
Lab work and activities help biology graduate Deidrie Briggs prepare for a career in optometry.
Our puzzler for this issue deals with probability. Suppose that a bin contains 12 balls – 5 red, 4 blue, and 3 yellow. We draw 3 balls at random from the bin, and we “win” this game if the three balls we choose represent exactly two colors. That is, we win if we choose two balls of the same color with the third ball being a different color. What is the probability of winning this game?
SolutionWe first need to know how many possible outcomes there are to this game. But that’s just a matter of some basic mathematics. At one time or another you probably met C(n,k), the number of ways to choose k objects out of a set of n objects. You might even remember the formula C(n,k) = n!/k!(n-k)!. Our game consists of choosing three balls from a set of 12, so the number of possible outcomes is C(12,3) = 12!/3!9! = (12*11*10)/(3*2) = 220.
Now all we need to do is figure out how many of those outcomes are “winning” outcomes. Well, it turns out that in this case it’s actually easier to count the number of outcomes that don’t win. To not win we would need to choose either all red, all blue, all yellow, or one ball of each color. Let’s take those one at a time:
- There are 5 red balls in the bin, so there are C(5,3) = 5!/3!2! = 10 ways to choose three of them.
- There are 4 blue balls in the bin, so there are C(4,3) = 4!/3!1! = 4 ways to choose three of them.
- There are only 3 yellow balls in the first place, so there’s only one way to choose all of them!
- There are 5*4*3 = 60 ways to choose one ball of each color.
- Jay Hunter (Chemistry, 1973)
- Daniel Schaal (Ph.D. Mathematics, 1994)
- Jana Stenback (Chemistry, 1986; Botany, 1994)