Probability, Part 1

So as I have acquired several new business statistics students, I thought I’d switch gears and talk about statistics. The first topic in statistics to get acquainted with is probability.

So what does it mean when the weatherperson says there’s an 80% chance of rain today? This is a probability expressed as a percentage. It means that given the weather conditions of today, 80 out of a hundred days like this will experience rain (on average). I say “on average” because a probability is just an indication of the likelihood of something occurring. Unless the probability is 0 (the event just will not happen) or 1 (the event will happen for sure), then the probability just expresses a likelihood of something happening. By the way, a probability is technically a number between 0 and 1. The 80% probability is the percentage equivalent to 0.80.

So most people intuitively know that the chance of flipping a head in a coin toss is 0.50 (50%). That does not mean that if you flip two coins, you only get one heads and one tails. But if you did this experiment 100 times, and counted the times you only get 1 heads, the average would tend to be 50, half or 50% of 100.

So how do you calculate a probability? The classic definition is

Probability = $\frac{{\mathrm{Number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{favorable}}\hspace{0.33em}{\mathrm{outcomes}}}{{\mathrm{Total}}\hspace{0.33em}{\mathrm{number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{possible}}\hspace{0.33em}{\mathrm{outcome}}{s}}$

So if we are interested in the probability of a heads in a single coin toss, the answer is

$\frac{{\mathrm{One}}\hspace{0.33em}{\mathrm{favorable}}\hspace{0.33em}{\mathrm{outcome}}\hspace{0.33em}{(}{\mathrm{h}}{\mathrm{e}}{\mathrm{a}}{\mathrm{d}}{\mathrm{s}}{)}}{{\mathrm{Possible}}\hspace{0.33em}{\mathrm{outcomes}}\hspace{0.33em}{(}{\mathrm{h}}{\mathrm{e}}{\mathrm{a}}{\mathrm{d}}{\mathrm{s}}\hspace{0.33em}{\mathrm{or}}\hspace{0.33em}{\mathrm{t}}{\mathrm{a}}{\mathrm{i}}{\mathrm{l}}{\mathrm{s}}{)}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{50}$

as expected.

What about the probability of getting a 1 on the roll of a fair die? Well the number of favorable outcomes is 1 and the total number of possibilities is 6. So the probability is $\frac{1}{6}$ which you can leave as a fraction.

What about the probability of getting an odd number in a singe throw of a die? The number of favorable outcomes is 3 (1, 3 or 5) and the total number of outcomes is 6. So the probability is $\frac{3}{6}$ or $\frac{1}{2}$ if you simplify this fraction. And it makes sense that on average, half the time when you throw a die, the number is odd and half the time the number is even.

This highlights a concept that the probability of all mutually exclusive and collectively exhaustive  outcomes has to be 1 as something has to happen. Mutually exclusive events are events where no other event can occur. For example throwing an odd number means that throwing an even number cannot occur. Collectively exhaustive means that all of the possible events are included. For example, throwing an odd number or throwing an even number are collectively exhaustive as there is no other possibility.

What is the probability of randomly selecting a face card from a shuffled deck of cards (no jokers)? Well there are 12 face cards in a deck so the probability is  $\frac{12}{52}$ or $\frac{3}{13}$

So far, the examples have been simple experiments with simple outcomes that can be easily counted so that the probability fraction can be easily derived. The complication comes when it is not so easy to count these. I will expand on this in forthcoming posts.

Posted on Categories Pre-VCE, Statistics