Answer:
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:
Alot! So the ones with 5 like 15,25,35,45,55,65,75,85,95! Each one inbetween is 10 so it would be 80
Answer:
5. No, Triangle J K L is a Reflection, not a translation
6. Its 11 to the right, and 2 up
