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.