There are several approaches to determine the probability of an event. One of them is what we call the classical or theoretical probability. Mathematically, the theoretical probability is obtained by dividing the number of the elements of the set called event by the number of the elements of the sample space, i.e. P(E) = \frac{\mid E \mid}{\mid S \mid}. In the formula, E is the event, S is the sample space., and P(E) is the theoretical probability of the event E.

 

Sample space
The sample space is the set of all possible outcomes of a statistical experiment. Throughout this post, the sample space is denoted by S. Each element in a sample space is called a sample point.

 

Example 1
If a dice is rolled once, there are 6 possible outcomes. The sample space S consists of 6 sample points (or 6 outcomes).

We write it simply S = {1, 2, 3, 4, 5, 6}.

 

Example 2
If 2 dice are rolled, there are 36 possible outcomes. Hence S consists of 36 sample points.

or simply, S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.

 

Example 3
If 2 coins are flipped simultaneously once, there are 4 possible outcomes. The sample space is S = {HH, HT, TH, TT}. In the sample space, HT means that the heads comes up on the first coin and the tails comes up on the second. TH means that the tails comes up on the first coin and the heads comes up on the second. HH and TT mean that both coins show the heads and the tails, respectively.

 

Events
An event is a subset of a sample space.

 

Example 1E
Referring to Example 1, let E be the event that a side with more than 4 spots occurs. In this case, E = {5, 6}. Note that E is a subset of S, meaning that every element of E is an element of S.

 

Example 2E
Referring to Example 2, let E be the event that the sides with an equal number of spots occur. In this case, E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}. Verify that E is a subset of S, that is, each member of E is a member of S.

 

Example 3E
Referring to Example 3, let E be the event that the sides of different types occur. Then, E = {TH, HT}.

 

What are the probabilities of the events in the previous examples?

To calculate these, use the formula P(E) = \frac{\mid E \mid}{\mid S \mid}. Here, \mid E \mid means the number of the elements of E and \mid S \mid means the number of the sample points in S.

 

Example 1P
A dice will be rolled once. What is the probability of getting a side with more than 4 spots?

Answer
Referring to Example 1 and Example 1E, we have S = {1, 2, 3, 4, 5, 6} and E = {5, 6}.
It follows that \mid S \mid = 6$ and $\mid E \mid = 2.
As a consequence, P(E) = \frac{\mid E \mid}{\mid S \mid} = \frac{2}{6} = \frac{1}{3}.
So, the probability of getting a side with more than 4 spots is \frac{1}{3}.

 

Example 2P
Two dice will be rolled. Find the probability that the sides with an equal number of spots occur.

Answer
Referring to Example 2 and Example 2E, we have:
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
Therefore, \mid S \mid = 36$ and $\mid E \mid = 6.
Consequently, P(E) = \frac{\mid E \mid}{\mid S \mid} = \frac{6}{36} = \frac{1}{6}.
So, the probability that the sides with an equal number of spots occur is \frac{1}{6}.

 

Example 3P
In flipping a pair of coins, find the probability that the sides of different types occur.

Answer
Referring to Example 3 and Example 3E, we have S = {HH, HT, TH, TT} and E = {TH, HT}. If follows that \mid S \mid = 4 and \mid E \mid = 2. Then, P(E) = \frac{\mid E \mid}{\mid S \mid} = \frac{2}{4} = 0.5. So, the probability that the sides of different types occur is 0.5.

 

PROBLEMS

No. 1
In a rolling of two dice (namely, the red and black dice) simultaneously, find the probabilities of the events below.
a. A = a sum of 9 appears.
b. B = the side with 4 spots occur on the black dice
c. C = an odd number of spots occur on the red dice but an even number of spots occur on the black dice
d. D = the dice show different number of spots
e. E = the red dice shows more than 4 spots and the black dice shows not more than 3 spots

No. 2
In a tossing of 3 coins simultaneously, find the probability that:
a. exactly 1 heads appears.
b. more than 1 tails appear.
c. less than 3 heads appear
d. at least one tails appear
e. all the coins show the same type of face

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