Well if the number cube goes up to 6 like most number cubes, then the probability would be 3/6 because there are 3 numbers that you want to roll (2, 3, and 5) out of 6 total numbers (1-6). Then 3/6 can be further simplified to 1/2. So the answer should be 1/2.
Answer: if you have infinity points, which i will asume are the events, they cant have the same probability because then the probability will not be normalized, because in graph of prob vs variable, you will se infinite area under the curve if the probability is constant.
And yes, can all points have positive probability of occurring, but besides you medium value (the bell for example) you will see an asintotic decrease to the zero.
Answer:
<em>The fraction of the beads that are red is</em>
Step-by-step explanation:
<u>Algebraic Expressions</u>
A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:
r:y = 2:3
y:b = 5:4
We are required to find r:s, where s is the total of beads in the bag, or
s = r + y + b
Thus, we need to calculate:
![\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Br%2By%2Bb%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B1%5D)
Knowing that:
![\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7By%7D%3D%5Cfrac%7B2%7D%7B3%7D%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B2%5D)
Multiplying the equations above:
Simplifying:
![\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Bb%7D%3D%5Cfrac%7B5%7D%7B6%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B3%5D)
Dividing [1] by r:
Substituting from [2] and [3]:
Operating:
The fraction of the beads that are red is 
