Understanding Probability
There are activities or experiments that do not have a fixed predictable outcome.
For example, tossing of a coin, rolling of a die, or drawing a card from a face down pack of cards. In such cases, although we cannot predict the exact result, we can predict the chances of one particular event happening.
For example, in the toss of a coin we can predict that we will get a head or a tail with equal chances.
The concept of assigning a number value to the chance of a particular outcome in an experiment is called $\underline{probability}$.
II. Values In Probability
In computing probability we assign a positive, real number between $0$ and $1$ to a particular event. It can be expressed as a fraction, decimal or percentage value.
Let us look at the meaning of probability values:
- A probability of $0$ means this event $cannot\ occur$. For example the probability of getting a sum of $1$ in the roll of two dice is $0$. Because the minimum sum that you can get in a roll of two dice is $2$, and it is impossible to get $1$ as the sum.
- A probability of $1$ means this event will always occur. For example, the probability of drawing a red ball from a box containing only red balls is $1$. Because no matter how many times you perform the experiment, you are sure to get a red ball as an outcome.
- The sum of probabilities of all possible events is always one. For example, the sum of probabilities of getting a $1$, $2$, $3$, $4$, $5$ or $6$ is always $1$. That is because, no matter what, your experiment will always lead to one of these values, and cannot lead to any other outcome.
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Probability value is the number obtained by dividing the number of favourable outcomes by the number of possible outcomes.
For example, what is the probability of getting a queen when you draw a card from a pack of $52$ cards?
The possible number of outcomes is $52$ out of which the favourable number of outcomes is $4$ (there are $4$ queens in a standard deck of cards).
Therefore, the probability of getting a queen is
$\dfrac{4}{52} = \dfrac{1}{13}$
Similarly, what is the probability of getting an odd number in the roll of a fair die?
The number of possible outcomes is $6$ out of which $3$ ($1$, $3$ and $5$) are odd, hence, are favourable outcomes.
Therefore, the probability is
$\dfrac{3}{6} = \dfrac{1}{2}$
.
A set of all possible outcomes of an experiment is called the sample space.
The sample space for a single roll of die will contain $6$ elements, which are as follows:
$\{1,\ 2,\ 3,\ 4,\ 5,\ 6\}$
The sample space of $3$ coin tosses will contain all of $8$ possible outcomes, which are:
$\{(H,\ H,\ H),\ (H,\ H,\ T),\ (H,\ T,\ H),\ (H,\ T,\ T),$$\ (T,\ H,\ H),\ (T,\ T,\ H),\ (T,\ H,\ T),\ (T,\ T,\ T)\}$
IV. Dependent & Independent Events
When you are conducting multiple runs of the same experiment, it is sometimes possible that the result of a particular run of the experiment is dependent on the previous experiment. These are called dependent events.
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For example, you are drawing balls from a box containing $3$ red and $3$ blue balls without replacing them. What is the probability of getting a blue ball in the second draw?
In the first draw, it is easy to see that the probability of getting a blue ball is $\dfrac{1}{2}$, since there are an equal number of blue and red balls. But during the second draw the outcome is dependent on what you got in the first round.
If the result of your first experiment was a blue ball, then the number of blue balls in the bag is $2$, and the total is $5$, and the probability of getting a blue ball in the second draw will be $\dfrac{2}{5}$ and if the first result was a red ball then the number of blue balls in the box will still be $3$ and the probability of getting a blue ball in the second draw would be $\dfrac{3}{5}$
In this case we can see that the outcome of the second experiment is clearly dependent on the first one. Hence, these are dependent events.
Now, let us see an example of independent events.
What is the probability of getting a head in the second toss of two consecutive tosses of a coin?
No matter what you obtained during the first toss, during the second toss there are still two equally likely outcome of head or tail. Hence, the probability of getting a head in the second toss is also $\dfrac{1}{2}$
V. Computing Simple Probabilities
There are two major cases that we will see here.
For any experiment, the probability of event $A$ happening is denoted by $P(A)$. The complement of event $A$ is usually denoted by $A'$ or $\overline{A}$, and its probability is denoted by $P(A')$ or $P(\overline{A})$.
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In any experiment, we can say for sure, that either event $A$ will happen or event $A$ will not happen. Like, in a roll of a die, you can either get a $5$ or $not$ get a $5$. One of these two $has$ to happen.
Therefore, $P(A) + P(A') = 1$
In case of $n$ independent experiments if the probability of outcome $A$ is $P(A)$ in one experiment, then the probability of getting $A$ as the outcome in $ALL$ the $n$ experiments is $P(A) \times P(A)... n\ times = \{P(A)\}^n$
For example, a single roll of a fair die, the probability of getting a $4$ is $\dfrac{1}{6}$.
Therefore, if $3$ dice are rolled, then the probability of getting all $4$s is
$\left(\dfrac{1}{6}\right)^3 = \dfrac{1}{6^3} = \dfrac{1}{216}$
.
VI. Mutually Exclusive Events
When two possible outcomes of an experiment, has no overlap or anything in common, they are called mutually exclusive events.
Let us consider the experiment, where you are drawing a ball from a bag containing red, blue and green coloured balls and we define the following events:
Event $A$ - You get a blue ball.
Event $B$ - You get a red or a green ball.
In this case the two events are mutually exclusive. But if we define the events as:
Event $C$ - You get a blue or a red ball
Event $D$ - You get a blue or a green ball
In this case, the events $C$ and $D$ have an overlap, and they are $NOT$ mutually exclusive.
In case of mutually exclusive events, it is true that the probability of either of them being the outcome, is the sum of their individual probabilities.
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In our example, probability of either $A$ or $B$ occurring is $P(A) + P(B)$, since they are mutually exclusive,
but the probability of either $C$ or $D$ occurring is $NOT$ equal to $P(C) + P(D)$ since they are not mutually exclusive.
We will take an example with real numbers:
In a roll of a die, we define the following events:
Event $A$ - The result is an even number
Event $B$ - The result is an odd prime number
Event $C$ - The result is a divisible by $3$
The total number of possible outcomes is $6$.
We can calculate the individual probabilities as:
$P(A) =$
$\dfrac{3}{6} = \dfrac{1}{2}$
(there are $3$ even numbers, they being, $2$, $4$ and $6$)
$P(B) =$
$\dfrac{2}{6} = \dfrac{1}{3}$
(there are $2$ odd prime numbers, which are: $3$ and $5$)
$P(C) =$
$\dfrac{2}{6} = \dfrac{1}{3}$
(the numbers $3$ and $6$ are divisible by $3$)
Now we can see that $A$ and $B$ are mutually exclusive whereas, $A$ and $C$ are not mutually exclusive since the outcome $6$ occurs in both.
We can say that the probability of getting an even number or an odd prime number $= P(A) + P(B) = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$
We can verify this by counting the number of favourable outcomes which are $2$, $4$, $6$, $3$ and $5$.
Now, if we look at $A$ or $C$ that is, the probability of getting an even number or a number divisible by $3$, the possible favorable events are $2$, $3$, $4$, $6$.
Therefore, the probability of getting $A$ or $C$ is
$\dfrac{4}{6} = \dfrac{2}{3}$
.
We can see that $P(A) + P(B) = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{5}{5} \ne \dfrac{2}{3}$
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Now, if we say that the probability of getting an even number or a number divisible by $6$ is $P(A) + P(C) = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ we will get an $INCORRECT$ result.
Let us verify. The set of favourable outcomes in this case would be $2$, $3$, $4$ and $6$
Therefore, the probability should be $\dfrac{4}{6} = \dfrac{2}{3}$