Chapter 5 The Normal Distribution and Some Basics of Probability

5.2.1 Working with Probabilities

We express probabilities as proportions (and we also denote them with p, just like we do proportions[1]), as this is indeed what they are:

 

    \[p=\frac{\textrm{number of specific outcomes we are interested in}}{\textrm{number of all possible outcomes}}\]

 

Or, the probability of a specific outcome is the proportion of the number of such outcomes out of the number of all possible outcomes.

 

Thus the probability of getting heads in a coin toss is:

 

    \[p(\textrm{heads})=\frac{\textrm{number of heads sides of a coin}}{\textrm{number of all sides of a coin}}=\frac{1}{2}=0.5\]

 

The same of course applies to tails:

 

    \[p(\textrm{tails})=\frac{\textrm{number of tails sides of a coin}}{\textrm{number of all sides of a coin}}=\frac{1}{2}=0.5\]

 

Heads and tails together exhaust all possible outcomes, so the probability that a coin will fall on any of its two sides is:

 

    \[p(\textrm{heads or tails})=\frac{2}{2}=\frac{1}{2}+\frac{1}{2}=0.5+0.5=1\]

 

Now how about we extend our example to something that has more that two outcomes? With six sides, a conventional die will serve us perfectly.

 

Following the same logic as with the coin, the probability to throw, say, a five is:

 

    \[p(\textrm{five})=\frac{\textrm{number of "five" sides of a die}}{\textrm{number of all sides of a die}}=\frac{1}{6}=0.167\]

 

The same goes for throwing a one, a two, a three, a four, or a six:

 

    \[p(\textrm{one})=p(\textrm{two})=p(\textrm{three})=p(\textrm{four})=p(\textrm{five})=p(\textrm{six})=\frac{1}{6}=0.167\]

 

Or, imagine you have a bowl with ten balls inside (i.e., the balls have numbers from 1 to 10). The probability of selecting each one out (without looking!) is, you guessed it, 1 out of 10, as each number appears only once and there are ten possible outcomes:

 

    \[p(1)=p(2)=p(3)=\ldots=p(10)=\frac{1}{10}=0.1\]

 

While this principle applies to N of any size — so we can increase the number of outcomes as much as we want — note the key prerequisite for the calculations to work: the outcomes must happen randomly. A coin toss and a die throw are classical examples of random chance. But when picking balls out of a bowl we have to make sure we don’t look or we might (consciously or subconsciously) choose one. Choosing a ball with a specific number introduces bias and thus invalidates randomness — i.e., it invalidates the principle of the outcomes having the same probability. Without this principle we cannot calculate anything: the only way to know the probability of an outcome is, in a sense, to divide the total probability, as it were, (i.e., 1) by the number of all possible outcomes, giving us equal probability for each. We know the probability of an outcome only if we know how many outcomes are possible in total and they all have the same probability. (Chapter 6 has more on the topic as it’s devoted to the topic of how random selection works.)

 


  1. If you need a reminder, the relevant part is in Section 2.3.1, here: https://pressbooks.bccampus.ca/simplestats/chapter/2-3-1-adding-percentages/

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