Let's dive into the basic terms on probability in brief.
Basic terms:
1. Random experiment or trial.
2. Outcome.
3. Sample point.
4. Sample space.
5. Event.
We have studied the above basics in the previous class. Click here to recall.
Probability of an event:
Shalini received 9 gifts on her birthday. Each gift is named from \(1\) to 9. When she is unpacking at random, what is the probability of picking gift box number \(3\)?
The word random tells us that each of the boxes have an equal chance of being picked up.
When a group of objects are mixed up, and a single object has to be picked up from the lot, then all of the objects have an equal chance of getting picked up at the first attempt. This makes the probability equally likely.
Probability of an event - Equally likely:
When a set of objects are mixed, and one object has to be picked up at random, then all the objects have an equal probability of getting picked up on the first attempt.
For an event \(E\), the probability of getting a favourable outcome is given by:
The probability of event \(E\) is shown as \(P(E)\).
\(P(E) =\) \(\frac{\text{The number of favourable outcomes}}{\text{The number of elements in the sample space}}\)
Continuing with the situation discussed earlier:
We know that 9 gifts are named as \(1\) to 9 is placed inside the bag.
Let the sample space, \(S\) be the collection of all possible outcomes.
Therefore, sample space, \(S = {1,2,3,4,5,6,7, 8, 9}\).
Thus, \(n (S) = 9\).
To find the probability of getting a \(3\):
Since only gift box number \(3\) is needed, the number of favourable outcomes is \(1\).
Thus, \(P (E) = \frac{1}{9}\).
What do you think the probability is of picking any \(1\) number at random?
We need any \(1\) number. Therefore, \(n (E) = 1\).
We know that, \(n (S) = 9\).
Thus, \(P (E) = \frac{1}{9}\).
Thus, the probability of picking any object of the mix at random is equally likely.
Complimentary events and probability
We denote the event that is 'not \(E\)' by E¯ . This is called the complement event of event \(E\).
So, \(P(E) + P(\text{not E}) = 1\).
We can rewrite it as, P(E)+P(E¯)=1, which gives us,  P(E¯)=1P(E).
In general, it is true that, for an event \(E\), P(E¯)=1P(E).
In a vegetable shop, there are \(5\) apples and \(7\) oranges in the basket. Shreya bought only \(5\) apples.
In this case, the number of apples Shreya bought can be taken as \(n(E)\), and the rest of the fruits in the basket, which she did not buy, taken as \(n\)\((\)E¯\()\).
Therefore, n(E)+n(E¯)=12, which gives us n(E¯)=12n(E).
Then, the probability of picking the apples in the basket \(P(E)\).
Similarly, the probability of picking the fruit other than apple (Orange) in the basket P(E¯).
Hence, P(E)+P(E¯)= \(\text{Total probability} = 1\).
i) The probability of each event lies between \(0\) and \(1\).
ii) The sum of all the probabilities is \(1\).
iii) \(E_1\), \(E_2\), . . .\(E_n\) covers all the possible outcomes of a trial.