Answer:
Probability of rolling at least 4 sixes is 0.01696.
Step-by-step explanation:
We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.
The above situation can be represented through binomial distribution;

where, n = number trials (samples) taken = 6 trials
r = number of success = at least 4
p = probability of success which in our question is probability of
rolling a “six", i.e; p = 0.20
<u><em>Let X = Number of sixes on a die</em></u>
So, X ~ Binom(n = 6, p = 0.20)
Now, Probability of rolling at least 4 sixes is given by = P(X
4)
P(X
4) = P(X = 4) + P(X = 5) + P(X = 6)
= 
= 
= 0.0154 + 0.00154 + 0.000064
= 0.01696
<em />
Therefore, probability of rolling at least 4 sixes is 0.01696.