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#### Example – 01:

- Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random what is the probability of getting a ticket bearing
**Solution:**

The sample space is S = {1, 2, 3, …….., 20}.

One ticket is drawn at random.

Hence n(S) = ^{20}C_{1} = 20

**an even number**

Let A be the event of getting ticket bearing an even number

Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20

∴ n(A) = ^{10}C_{1} = 10

By the definition P(A) = n(A)/n(S) = 10/20 = 1/2

Therefore the probability of getting ticket bearing even number is 1/2

**an odd number**

Let B be the event of getting ticket bearing an odd number

Favourable points are , 3, 5, 7, 9, 11, 13, 15, 17, 19

∴ n(B) = ^{10}C_{1} = 10

By the definition P(B) = n(B)/n(S) = 10/20 = 1/2

Therefore the probability of getting ticket bearing odd number is 1/2

**a perfect square**

Let C be the event of getting ticket bearing a perfect square

Favourable points are 1, 4, 9,16

∴ n(C) = ^{4}C_{1} = 4

By the definition P(C) = n(C)/n(S) = 4/20 = 1/5

Therefore the probability of getting ticket bearing a perfect square is 1/5

**multiple of four (or divisible by four)**

Let D be the event of getting ticket bearing a number multiple of 4

Favourable points are 4, 8, 12, 16, 20

∴ n(D) = ^{5}C_{1} = 5

By the definition P(D) = n(D)/n(S) = 5/20 = 1/4

Therefore the probability of getting ticket bearing a number multiple of 4 is 1/4

**multiple of three (or divisible by three)**

Let E be the event of getting ticket bearing a number multiple of 3

Favourable points are 3, 6, 8, 12, 15, 18

∴ n(E) = ^{6}C_{1} = 6

By the definition P(E) = n(E)/n(S) = 6/20 = 3/10

Therefore the probability of getting ticket bearing a number multiple of 3 is 3/10

**a number greater than 4**

Let F be the event of getting ticket bearing a number greater than 4

Favourable points are 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

∴ n(F) = ^{17}C_{1} = 17

By the definition P(F) = n(F)/n(S) = 17/20

Therefore the probability of getting ticket bearing a number greater than 4 is 17/20

**a number less than 11**

Let G be the event of getting ticket bearing a number less than 11

Favourable points are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

∴ n(G) = ^{10}C_{1} = 10

By the definition P(G) = n(G)/n(S) = 10/20 = 1/2

Therefore the probability of getting ticket bearing a number less than 11 is 1/2

**perfect square or less than 5**

Let H be the event of getting ticket bearing a number a perfect square or less than 5

Favourable points are 1, 4, 9, 16, 2, 3

∴ n(H) = ^{6}C_{1} = 6

By the definition P(H) = n(H)/n(S) = 6/20 = 3/10

Therefore the probability of getting ticket bearing a perfect square or less than 5 is 3/10

**a prime number**

Let J be the event of getting ticket bearing a number a prime number

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19

∴ n(J) = ^{8}C_{1} = 8

By the definition P(J) = n(J)/n(S) = 8/20 = 2/5

Therefore the probability of getting ticket bearing a prime number is 2/5

**a prime number or a perfect square**

Let K be the event of getting ticket bearing a number a prime number or a perfect square

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19, 1, 4, 9, 16

∴ n(K) = ^{12}C_{1} = 12

By the definition P(K) = n(K)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

**an even number or a perfect square**

Let L be the event of getting ticket bearing an even number or a perfect square

Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20, 1, 9

∴ n(L) = ^{12}C_{1} = 12

By the definition P(L) = n(L)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

**an even number or a number divisible by 5**

Let M be the event of getting ticket bearing an even number or a number divisible by 5

Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20, 5, 15

∴ n(M) = ^{12}C_{1} = 12

By the definition P(M) = n(M)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

**a perfect square or a number multiple of 3**

Let N be the event of getting ticket bearing a perfect square or a number multiple of 3

Favourable points are 1, 4, 9, 16, 3, 6, 12,15, 18

∴ n(N) = ^{9}C_{1} = 9

By the definition P(N) = n(N)/n(S) = 9/20

Therefore the probability of getting ticket bearing a perfect square or a number multiple of 3 is 9/20

**a number greater than 9 and an even number.**

Let Q be the event of getting ticket bearing a number greater than 9 and an even number

Favourable points are 10, 12, 14, 16, 18, 20

∴ n(Q) = ^{6}C_{1} = 6

By the definition P(Q) = n(Q)/n(S) = 6/20 = 3/10

Therefore the probability of getting ticket bearing a number greater than 9 and an even number is 3/10

**an even number and multiple of 3**

Let R be the event of getting ticket bearing an even number and multiple of 3

Favourable points are 6, 12, 18

∴ n(R) = ^{3}C_{1} = 3

By the definition P(R) = n(R)/n(S) = 12/20 = 3/5

Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5

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