Relation between Empirical and Theoretical Probability

Theoretical Probability is a term used to find the chances of occurrence of a particular event. For example, the outcome of tossing a coin produces either heads or tails.

Empirical Probability on the other hand is more of an approximation. It gives us the number of times an event is expected to occur, knowing its nature and previous occurrences. Thus, the empirical probability is not an experimental probability. It is simply determined analytically, i.e. by using the knowledge about the nature of the event. Hence, it is also known as Estimated Probability. For example, tossing a fair coin and observing its uppermost side. Now, we do not know the outcome of the event, but we do know the nature of the coin. Thus, we can predict the probability as:

If S is the focus, ZZ1 is the directrix and P is any point on the ellipse, then the definition

P (Tail) = ½, and P (Head) = ½

P (E) = number of times event E occurs / total number of observed occurrences

Theoretical Probability in Which All the Events Are Equally Likely

In an experiment where all the outcomes are equally likely (example, tossing of a coin, occurrence of a number on the throw of a die, etc), the empirical probability is calculated by the formula

P (E) = Number of favorable outcomes in E / Total number of outcomes

P(E) = n(E) / n(S)

Example 6

A boy has two red and two yellow balls, making it a total of 4 balls.

        

Calculate the empirical probabilities of all the cases, given that the boy selects only two out of the four balls.

Solution:

Note that the probability of the second draw is affected by the outcome of the first draw

Case I: Both the balls are red.

  

P (both red) = P (first ball is red) x P (second ball is red)

P (first ball is red) = 2/4 = ½

P (second ball is red) = 1/3

Therefore P (both red) = 1/6

Case II: The first ball is yellow, and the second is red.

  

P (yellow, red) = ½

P (first ball is yellow) = ½

P (second ball is red) = 2/3

Therefore P (first yellow, second red) = 2/6 = 1/3

Case III: Both the balls are yellow

  

P (both yellow) = 1/6, as per P (both red) above

Case IV: The first ball is red, and the second is yellow

  

P (red, yellow) = 1/3, as per P (first yellow, second red) above

Note that the sum of all probabilities (that is ALL possible occurrences must equal 1)

Try this question

1. In a football game, a player hits 3 goals out of 10 attempts. What is the empirical probability that the next shot will be a goal?
   1. 1
   2. 0
   3. 3/10
   4. Data is insufficient
   Answer: C
   Favorable outcomes=3
   Total attempts=10
   Probability=3/10
2. A pair of dice is rolled. Find the probability that the sum of the two numbers that appear in each of the two die is 7.
   1. 1/7
   2. 1/6
   3. 1/5
   4. 1/4
   Answer: B
   Total outcomes, n(S)=6.6 =36
   Favorable outcomes, n(E)= {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}=6
   Thus, probability= n(E)/ n(S)
   P=1/6
3. In a bowling game, Johnny hits 4 out of 10 pins in the first frame. Each frame gives a player 2 chances to knock down all 10 pins. What is the empirical probability that the next frame he will hit all pins?
   1. 14/10
   2. 2/5
   3. 6/10
   4. 4/6
   Answer: B
   1 frame = 10 pins
   Favorable outcomes = 4
   Total attempts = 10
   Probability = 4/10
   = 2/5