A fair die is cast four times. Calculate
the probability of obtaining at least two
S6's.
Round to the nearest tenth of a
percent.

I'd say that is an "occupancy problem".

I ran a spreadsheet simulation of that and I'd say the probability is approximately .13

Those problems are rather complex to solve.  What I think you would have to do is calculate the probability of  

A) ZERO sixes appearing in 4 rolls.

B) exactly 1 six appears in 4 rolls.

C) exactly 2 sixes appear in 4 rolls.

D) exactly 3 sixes appear in 4 rolls. and

E) exactly 4 sixes appear in 4 rolls.

4 rolls of a die can produce 6^4 or 1,296 combinations.

A) is rather easy to calculate:  The probability of NOT rolling a six in one roll is 5/6.  In 4 rolls it would be (5/6)^4 = 0.4822530864

E) is fairly easy to calculate: The probability of rolling one six is (1/6).  The probability of rolling 4 sixes is (1/6)^4 = 0.0007716049

Then we need to:

D) calculate how many ways can we place 3 objects into 4 bins

C) calculate how many ways can we place 2 objects into 4 bins

B)  calculate how many ways can we place 1 objects into 4 bins

I don't know how to calculate D C and B

Step-by-step explanation:

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8