Three dice are thrown. The probability of getting a sum of numbers which is a perfect cube, is: |
$\frac{1}{6}$ $\frac{1}{8}$ $\frac{5}{36}$ $\frac{7}{72}$ |
$\frac{7}{72}$ |
If three dice are thrown together, sample space = {(111), (112).....(166) (211),...............(266) (311),...............(366) (411),...............(466) (511),...............(566) (611),...............(666)} Total number of outcomes = 216 Minimum sum of the three outcomes = 3 Maximum sum of the three outcomes = 18 Perfect cubes in 3 to 18 = 8 Chances of getting the sum as 8 = (116)(161)(611)(125)(152)(215)(251)(521)(512)(134)(143)(413)(431)(314)(341)(224)(242)(422)(332)(232)(233) Required probability = 21/216 = 7/72 The correct answer is Option (4) → $\frac{7}{72}$ |