Answer:
Explanation:
The sample space in a 3 times die roll =63=216. 
 
Now, P(getting at least one 6)=1−P(getting NO 6). 
 
Probability of getting no 6 in a single die roll 
 
=56. 
 
Thus probability of getting NO 6 in three die rolls 
 
=56×56×56=125216. 
 
Therefore, the probability of getting at least one 6 
 
P=1−125216=91216.