Browse Questions

Three identical dice are rolled. The probability that the same number will appear on each of them is

$\begin {array} {1 1} (A)\;\large\frac{1}{6} & \quad (B)\;\large\frac{1}{36} \\ (C)\;\large\frac{1}{18} & \quad (D)\;\large\frac{3}{28} \end {array}$

Three dice can fall in - 216 ways
If the same number appears on each of them then the possible number of ways is 6.
Hence the required probability is $\large\frac{6}{216}$
$= \large\frac{1}{36}$
And : (B)
but question has said that three IDENTICAL dices are thrown
so (1,2,3) is same as (3,2,1) so answer should be 3/28