Question

Two dice are thrown together. The probability that at least one will show its digit greater than 3 is

विकल्प

• 1/4

•  3/4

•  1/2

• 1/8

   

Answer 1

 3/4

When two dice are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space, given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
      (2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
      (3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
      (4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
      (5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
      (6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
i.e. n(S) = 36
Let E be the event of getting at least one digit greater than 3.
Then E = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
          (5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
∴  n(E) = 27
Hence, required probability = \[\frac{27}{36} = \frac{3}{4}\]