Mathematical Probability (#1)

In daily life, people often interchange the words ‘possibility’ and ‘probability’. However, these words are NOT interchangeable. “Possibility” is used when there is something that is said in qualitative terms (mostly as a vague statement). It is related to most of the tasks, which are assessed to determine whether or not they are possible to do in a given set of circumstances.

On the other hand, “probability” is a quantitative measurement of an event or a set of all the possible events with respect to an object or a system of objects.

In all the existing systems, the mathematical probability always ranges from 0 to 1. It can be any real number between 0 and 1, with 0 meaning that a particular event is ‘improbable’ (meaning such an event does not exist) and 1 meaning that the event will certainly occur (a 100% probable event).

Let us explore probability with the help of a few examples.

  1. A simple unbiased coin:
    • For a simple unbiased coin, there are only two events that can be observed upon a toss. Heads (denoted by H), and Tails (denoted by T). These two possible events form the sample space. A sample space is the number of all the possible events for a certain system. For the coin in consideration, the numerical value of the sample space is 2.
    • Therefore, the probability of either event happening for a coin is as follows:
    •  P_{(Heads)} = \frac{1}{2}
    •  P_{(Tails)} = \frac{1}{2}
  2. Two coins tossed simultaneously:
    • If we toss two coins simultaneously, there are two events possible for each coin, independent of the other coin.
    • Hence, the events that can occur for such a system are as follows.
Coin ACoin B
Event AHeadsHeads
Event BHeadsTails
Event CTailsHeads
Event DTailsTails

If carefully observed, for one coin tossed, the sample space was 2. For two coins tossed, the sample space was 4, which is  \ 2^2 .

Similarly, it can be easily concluded that for n coins tossed simultaneously, the sample space will be  \ 2^n . Let’s take the example of 2 coins tossed to prove this. For coin A Heads, there are two events for coin B, and for coin A Tails, there are two possible events, as seen in the table above. Hence, for every first coin heads, there will be  \ 2^{(n-1)} possible events that occur, and similarly for tails. Hence, the sample space is  2 \times 2^{(n-1)} = 2^n .

Now, let’s take the example of an unbiased die. An unbiased die has six faces, each with a numerical value on it like 1, 2, 3, 4, 5, or 6 dots. When the die is thrown, one of the six faces will be on the upper side. Therefore, the sample space is 6, and the probability for each event is as follows:
$$ P_{\text{(1 dot will face up)}} = \frac{1}{6} $$
$$ P_{\text{(2 dots will face up)}} = \frac{1}{6} $$
$$ P_{\text{(3 dots will face up)}} = \frac{1}{6} $$
$$ P_{\text{(4 dots will face up)}} = \frac{1}{6} $$
$$ P_{\text{(5 dots will face up)}} = \frac{1}{6} $$
$$ P_{\text{(6 dots will face up)}} = \frac{1}{6} $$

Similarly, like the multiple coin toss, if two dice are thrown simultaneously, for each face up from one die, there are six possible outcomes from the other die. Hence, the sample space is  6^2 . Also, for n dice thrown simultaneously, the sample space will be  6^n .
{Note: The probability can never be less than zero. And the sum of all probabilities must be equal to one.}

To generalize this, if we have a die of ‘m’ sides, and ‘n’ such dice are thrown simultaneously, the sample space will be  m^n .

As a final example for this chapter, I will consider the example of the deck of 52 cards (not considering jokers and any other miscellaneous cards).
An unbiased deck of cards has the following:
13 cards from “Hearts” group (Red)
13 cards from “Spades” group (Black)
13 cards from “Clubs” group (Black)
13 cards from “Diamond” group (Red)

Also, there are 4 cards for each number or letter (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, and A).
Hence, the sample space for the deck of cards is 52.

  1. If a card drawn at random is a 4 and red, then its probability is  \frac{2}{52} = \frac{1}{26} , since there are two red cards that have the value 4 and are red.
  2. If a card drawn at random is a numbered red card, then its probability is  \frac{9 \times 2}{52} = \frac{18}{52} = \frac{9}{26} , since there are two red suits and each suit has 9 cards that have numbers on them.

Looking at the above examples, we can understand that to find the mathematical probability, we need to find out the ratio between the numerical value of the event in consideration and the numerical value of the sample space ( \frac{\text{numerical value of occurrence of specific event}}{\text{numerical value of sample space}} ). However, it is best when the student has knowledge of exponents and fractions at this stage to navigate through this topic with ease.

Once the student starts practicing the field and methods of finding the probability, they will find out that by finding all possible scenarios and weighing the probability of each one, we can understand many systems and determine our approach to solving various problems. All the best!


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