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:
Each side is 52 cm. because equilateral triangles have equal lengths on each side. <span />
Danny's score at the end of the game is 30 because:
We started our with 10 points.
He loses 20 points, which brings him down to -10 points.
In the final round he earned 40 points. 40 minus the negative ten points gives us an answer of 30 points.
Answer:
Step-by-step explanation:
<u>9</u>/14 × 5/<u>3</u>
The factors can be canceled if they are factors of both the numerator of the first fraction and the denominator of the second fraction. The factors get cancelled leaving the second fraction to a whole number.
3/14 × 5
(3 × 5)/14
15/14
