 UPSKILL MATH PLUS

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Empirical approach
When a probability is found considering the previously observed data or based on a larger set of observations, it is called an empirical probability approach.
Example:
When a $$20000$$ pairs of shoes are ordered from the company and $$2500$$ of them are faulty, what is the number of faulty pair of shoes?
The empirical probability is considered in the case of a large number of trials.

How to calculate the empirical probability:

Let $$P(E)$$ be the probability of an event. For a large number of trials, $$n$$, the number of probable outcomes is $$r$$.

In this case, $$P(E) = \frac{r}{n}$$.

We know that the probability will always be between 0 and 1.

This can also be written as $$0 \leq P(E) \leq 1$$.

Let us look at the construction of the above given expression in detail.

Firstly, we know that $$r$$ cannot be greater than $$n$$. In case $$r$$ is greater than $$n$$, the probability value becomes greater than $$1$$. It defies the primary rule of probability.

Therefore, $$P(E) =$$ $$\frac{r}{n}$$ $$< 1 \longrightarrow (1)$$

If $$r = 0$$:

$$P(E) =$$ $$\frac{0}{n}$$ $$= 0$$ $$\longrightarrow (2)$$

If $$r = n$$:

$$P(E) = \frac{n}{n} = 1 \longrightarrow (3)$$

From $$(1)$$, $$(2)$$, and $$(3)$$, it is clear that:

$$0 \leq P(E) \leq 1$$
Example:
Continuing with the situation discussed earlier:

The number of trials, $$n = 20000$$

The number of faulty pair of shoes, $$r = 2500$$

Then, $$P(E) = \frac{2500}{20000}$$

$$P(E) = 0.125$$