Probability: Problems Based on Throwing a Cubic Die

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Cubic die

  • 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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Next Topic: Problems Based on Throwing of Two Dice 

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