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- A fair die is thrown
- The sample space is S = {1, 2, 3, 4, 5, 6} and n(S) = 6

#### Example – 01:

- A perfect cubic die is thrown. Find the probability of that

A perfect cubic die is thrown

The sample space for the experiment is

S = {1, 2, 3, 4, 5, 6}

∴ n(S) = 6

**an even number comes up or a number divisible by 2 comes up**

Let A be the event of that an even number comes up

∴ A = {2, 4, 6}

∴ n(A) = 3

By the definition P(A) = n(A)/n(S) = 3/6 = 1/2

Therefore the probability that an even number comes up is 1/2

**an odd number comes up**

Let B be the event of that an odd number comes up

∴ B = {1, 3, 5}

∴ n(B) = 3

By the definition P(B) = n(B)/n(S) = 3/6 = 1/2

Therefore the probability that an odd number comes up is 1/2

**a number multiple of 3 comes up**

Let C be the event of that a number multiple of 3 comes up

∴ C = {3, 6}

∴ n(C) = 2

By the definition P(C) = n(C)/n(S) = 2/6 = 1/3

Therefore the probability that a number multiple of 3 comes up is 1/3

**a number multiple of 3 or 5 comes up**

Let D be the event of that a number multiple of 3 or 5 comes up

∴ D = {3, 6, 5}

∴ n(D) = 3

By the definition P(D) = n(D)/n(S) = 3/6 = 1/2

Therefore the probability that a number multiple of 3 or 5 comes up is 1/2

**a number multiple of 3 and 5 comes up**

Let E be the event of that a number multiple of 3 and 5 comes up

∴ E = {} = Φ

∴ n(E) = 0

By the definition P(E) = n(E)/n(S) = 0/6 = 0

Therefore the probability that a number multiple of 3 and 5 comes up is 0

**Note: Such events with zero probability are called impossible events**

**the score is greater than 2**

Let F be the event of that the score is greater than 2

∴ F = {3, 4, 5, 6}

∴ n(F) = 4

By the definition P(F) = n(F)/n(S) = 4/6 = 2/3

Therefore the probability that the score is greater than 2 is 2/3

**a perfect square comes up**

Let G be the event of that a perfect square comes up

∴ G = {1, 4}

∴ n(G) = 2

By the definition P(G) = n(G)/n(S) = 2/6 = 1/3

Therefore the probability that a perfect square comes up is 1/3

**A prime number comes up**

Let H be the event of that a prime number comes up

∴ H = {2, 3, 5}

∴ n(H) = 3

By the definition P(H) = n(H)/n(S) = 3/6 = 1/2

Therefore the probability that a prime number comes up is 1/2

**Note: 1 is not a prime number**

**the score is less than 5 but not less than 2**

Let J be the event of that score is less than 5 but not less than 2

∴ J = {3, 4}

∴ n(J) = 2

By the definition P(J) = n(J)/n(S) = 2/6 = 1/3

Therefore the probability that the score is less than 5 but not less than 2 is 1/3

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