### Theory:

**Consider the following situation**.

If 5 boxes named \(1\) to 5 is placed inside a bag, what is the probability of picking box number 5 at random?

The word random tells us that each of the boxes has 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 \(P(E)\).

The probability of event \(E\) is shown as \(P(E)\).

\(\frac{\text{The number of favourable outcomes}}{\text{The total number of possible outcomes}}\)

$P(E)=\frac{n(E)}{n(S)}$, where p is $0\le p\le 1$.

Example:

**Continuing with the situation discussed earlier**:

We know that 5 boxes named \(1\) to \(5\) is placed inside the bag.

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

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

Thus, \(n(S) = 5\)

**To find the probability of getting a**\(4\):

Since only box number \(4\) is needed, the number of favourable outcomes is \(1\).

Thus, \(P(E) =\) $\frac{1}{5}$

Important!

**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.