Probability: Problems Based on Numbered Tickets or Cards – 01

Physics Chemistry  Biology  Mathematics
Science > Mathematics > ProbabilityYou are Here

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) = 20C1 = 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) = 10C1 = 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) = 10C1 = 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) = 4C1 = 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) = 5C1 = 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) = 6C1 = 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) = 17C1 = 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) = 10C1 = 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) = 6C1 = 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) = 8C1 = 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) = 12C1 = 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) = 12C1 = 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) = 12C1 = 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) = 9C1 = 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) = 6C1 = 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) = 3C1 = 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

Previous Topic: Problems Based on Throwing of Two or More Dice

Next Topic: Problems Based on Selection of Two or More-Numbered Tickets

Science > Mathematics > ProbabilityYou are Here
Physics Chemistry  Biology  Mathematics

Leave a Comment

Your email address will not be published. Required fields are marked *