Probability – An Experimental Approach

Experimental probability is the result of probability based on the actual experiments.

1. It is also called Empirical Probability.
2. In this probability, the results could be different, every time you do the same experiment.
3. As the probability depends upon the number of trials and the number of times the required event happens.
4. If the total number of trials is ‘n’ then the probability of event E happening is

P(E) = $$\frac{Number of trials in which the event happend}{The total number of trials}$$

Example 1: A coin is tossed 1000 times with the following frequencies:

Head: 455, Tail: 545

compute the probability for each event.

sol: Since the coin is tossed 1000 times, the total number of trails is 1000. let us call the events of getting a head and of getting a tail as E and F, respectively. Then, the number of times E happens, i.e., the number of times a head come up, is 455.

So,   the probability of E = $$\frac{Number of heads}{Total number of trials}$$

i.e.,   P(E) = $$\frac{455}{1000}$$ = 0.455

similarly, the probabilyty of the event of getting a tail = $$\frac{Number of tails}{Total number of trials}$$

i.e.,    P(F) = $$\frac{545}{1000}$$ = 0.545

Note that in the example above P(E) + P(F) = 0.455 + 0.545 = 1, and E and F are the onlt two possible outcomes of each trail.

Example 2: A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table :

Find the probability of getting each outcome.

Solution : 

 Let E, denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.

Then

probability of outcome 1 = $$P(E_{1})$$ = $$\frac{Frequency of 1}{Total number of times the die is thrown}$$

= $$\frac{179}{1000}$$ = 0.179

Similarly, $$P(E_{2})$$ = $$\frac{150}{1000}$$ = 0.15, $$P(E_{3})$$ = $$\frac{157}{1000}$$ = 0.157,

$$P(E_{4})$$ = $$\frac{149}{1000}$$ = 0.149, $$P(E_{5})$$ = $$\frac{175}{1000}$$ = 0.175

and $$P(E_{6})$$ = $$\frac{190}{1000}$$ = 0.19

Note that $$P(E_{1})$$ + $$P(E_{2})$$ + $$P(E_{3})$$ + $$P(E_{4})$$ + $$P(E_{5})$$ + $$P(E_{6})$$ = 1