## Understanding Sample Space and Sample Points

In tossing a fair coin, there are only two possible outcomes, a Head (H) and a Tail (T).  If we let S be the set of all possible outcomes of this event, then, we write the set of possible outcomes as S = {H,T}.

If two fair coins are tossed, then the outcomes can be both heads {H,H} or both tails {T,T}. It can also be a head first then a tail {H,T}, or a tail first and then a head {T,H}. So, in tossing two coins, we have the set of possible outcomes S = {{T,T}, {T,H}, {H,T}, {H,H}}.

As the number of tosses increases, listing gets more difficult. One of the strategies that can be used to remedy this problem is by creating a tree diagram. The following problem is solved using a tree diagram. Notice that it is a three-coin tossing problem in disguise (try replacing B with H and G with T). » Read more

## Wedding Guests and Circular Permutations

In a wedding banquet, guests are seated in circular table for four. In how many ways can the guests be seated?

We have learned that the number of permutations of $n$ distinct objects on a straight line is $n!$. That is, if we seat the four guests Anna, Barbie, Christian, and Dorcas, on chairs in on a straight line they can be seated in $4 \times 3 \times 2 \times 1 = 24$ ways (see complete list).

However, circular arrangement is slightly different. Take the arrangement of guests A, B, C, D as shown in the first figure.  The four possible seating arrangements are just a single permutation: in each table, the persons on the left and on the right of each guest are still the same persons. For example, in any of the tables, B is on the left hand side of A and D is on the right hand side of A. In effect, the  four  linear permutations ABCD, BCDA, CDAB, and DABC are  as one in circular permutation. This means that the number of linear permutations of 4 persons is four times its number of circular permutations.  Since the number of  all possible permutations of four objects is 4!, the number of circular permutations of four objects is $\frac{4!}{4}$. » Read more

## Lottery Math – If You Play You Should Know Your Odds

Pop Quiz

Here’s a little pop quiz for you, my friends. Sorry, it’s a math quiz. But don’t worry, it’s easy and this is a math blog, after all. Little Johnny, a fifth grader, came home from school one day, visibly upset. His mommy asked what was wrong. Little Johnny told her that he lost out on a treat that his teacher had. He explained to his mommy that his teacher had just one treat, but there were ten students in the class. So each of the students put their names into a hat. Then the teacher randomly chose one name to receive the treat. The rest of the class got nothing. Little Johnny didn’t get the treat. He was upset.

To comfort him, mommy told Little Johnny that the odds of him getting the treat were against him from the start; his odds were a mere 1-in-___ (Fill in the blank).

Little Johnny didn’t appreciate his mommy spewing out odds as if she was a bookie. Seriously, is that all about, mommy? However, being a simple fifth grade math question, he knew the answer (as should you).

1 2 3