### Theory:

The classical approach is the oldest approach of calculating probability.

**Consider the following situation**.

If \(7\) boxes named \(1\) to \(7\) is placed inside a bag, what is the probability of picking box number \(3\) at random?

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.

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.

**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 \(P(E)\).

\(P(E) =\) \(\frac{\text{The number of favourable outcomes}}{\text{The number of elements in the sample space}}\)

\(P(E) =\) \(\frac{n(E)}{n(S)}\)

Example:

**Continuing with the situation discussed earlier**:

We know that \(7\) boxes named as \(1\) to \(7\) is placed inside the bag.

Let the sample space, \(S\) be the collection of all possible outcomes.

Therefore, sample space, \(S =\) \(\{1, 2, 3, 4, 5, 6, 7\}\).

Thus, \(n(S) = 7\)

Since only box number \(3\) is needed, the number of favourable outcomes is \(1\).

Thus, \(P(E) = \frac{1}{7}\)

We need any \(1\) number. Therefore, \(n(E) = 1\)

We know that, \(n(S) = 7\)

Thus, \(P(E) = \frac{1}{7}\)

Thus, the probability of picking any object of the mix at random is equally likely.

Thus, \(n(S) = 7\)

**To find the probability of getting a**\(3\):Since only box number \(3\) is needed, the number of favourable outcomes is \(1\).

Thus, \(P(E) = \frac{1}{7}\)

**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, \(n(S) = 7\)

Thus, \(P(E) = \frac{1}{7}\)

Thus, the probability of picking any object of the mix at random is equally likely.

Important!

**1.**The probability value would always be between \(0\) and \(1\).

**2.**The probability of equally likely outcomes are always equal.

**3.**If \((V)\) is the probability of success of an event, then:

The probability of failure, \(n(F)\), of the same event \(= 1 - n(V)\)