Question

Three symmetrical dice are thrown. The probability of having different points on them is

• A. $\frac{35}{36}$
• B. $\frac{4}{9}$
• C. $\frac{5}{9}$
• D. $\frac{1}{36}$

Answer 1

The correct option is C $\frac{5}{9}$

Probability of having different points on three dice $=$ 1 $-$ Probability of having same points on the dice
Cases when same points on the dice:
Case 1: When all points are same: 1 1 1, 2 2 2, 3 3 3, 4 4 4, 5 5 5, 6 6 6 $=6$ cases
Case 2: When two points are same and one is different:
$(11-)$: 5 and we could arrange on three ways so 15 cases are there
Similarly for $(22-);(33-);......(66-)$.
Total cases when two points are same and one is different: $15\times 6=90$
Total cases when same points on the dice: $6+90=96$
Probability of having different points on three dice $=1-\frac{96}{6\times 6\times 6}=\frac{5}{9}$