**Axiomatic definition of probability :**

Let us consider a sample space S in connection with a random experiment and let A be an event defined on the sample space S. That is, A ≤ S.

Then a real valued function P defined on S is known as a probability measure and P(A) is defined as the probability of A if P satisfies the following axioms :

**Problem 1 :**

A number is selected from the first 25 natural numbers. What is the probability that it would be divisible by 4 or 7 ?

**Solution :**

Let A be the event that the number selected would be divisible by 4 and B, the event that the selected number would be divisible by 7.

Then AuB denotes the event that the number would be divisible by 4 or 7.

Next we note that

A = {4, 8, 12, 16, 20, 24} and B = {7, 14, 21}

whereas S = {1, 2, 3, ……... 25}.

Here, AnB = Null set.

The two events A and B are mutually exclusive and as such we have

P(AuB) = P(A) + P(B) ----- (1)

Since

P(A) = n(A) / n(S) = 6/25

and

P(B) = n(B) / n(S) = 3/25

Thus from (1), we have

P(AuB) = 6/25 + 3/25

P(AuB) = 9/25

Hence the probability that the selected number would be divisible by 4 or 7 is 9/25 or 0.36

**Problem 2 :**

A coin is tossed thrice. What is the probability of getting 2 or more heads ?

**Solution :**

If a coin is tossed three times, then we have the following sample space.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

2 or more heads imply 2 or 3 heads.

Next we note that

If A and B denote the events of occurrence of 2 and 3 heads respectively, then we find that

A = { HHT, HTH, THH } and B = { HHH }

P(A) = n(A) / n(S) = 3 / 8

and

P(B) = n(B) / n(S) = 1 / 8

As A and B are mutually exclusive, the probability of getting 2 or more heads is

P(AuB) = P(A) + P(B)

P(AuB) = 3/8 + 1/8

P(AuB) = 0.50

Hence, the probability of getting 2 or more heads is 0.50

After having gone through the stuff given above, we hope that the students would have understood "Axiomatic definition of probability".

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