To begin with, consider a straightforward hidden Markov model (HMM). We observe a series of outcomes from rolling a four-sided die at an "occasionally dishonest casino". At time t, the result x_t belongs to the set {1, 2, 3, 4}. The casino can either be in state z_t belonging to {1, 2}. When z_t is equal to 1, it uses a fair die, whereas when z_t is equal to 2, the die is biased towards rolling a 1. Specifically: p (x_t = 1 | z_t = 1) = p (x_t = 2 | z_t = 1) = p (x_t = 3 | z_t = 1) = p (x_t = 4 | z_t = 1) = 0.25, p (x_t = 1 | z_t = 2) = 0.7, and p (x_t = 2 | z_t = 2) = p (x_t = 3 | z_t = 2) = p (x_t = 4 | z_t = 2) = 0.1. Assume there is an equal likelihood of starting in either state at time t = 1, which leads to p (z1 = 1) = p (z1 = 2) = 0.5. The casino generally maintains the same die for several iterations, but it occasionally switches states with these probabilities: p (z_t + 1 = 1 | z_t = 1) = 0.8 and p (z_t + 1 = 2 | z_t = 1) = 0.2; likewise, p (z_t + 1 = 2 | z_t = 2) = 0.1 and p (z_t + 1 = 1 | z_t = 2) = 0.9. To find the probability p (z1 = z2 = z3) that the same die is used across the first three rolls under the HMM generative model, consider the following. If we assume the first die is state 1, the probability can be calculated as p(z1=1)=0.5, and consequently, p(z2=1|z1=1)=0.8 signifies that the same die might still be in use. Alternatively, if we start with the die in state 2, p(z1=2)=0.5 and p(z2=2|z1=2)=0.9 also provides a probability. Adjacent transition probabilities can be expressed as follows: p(z_t+1=2|z_t=1)=1-p(z_t+1=1|z_t=1)=0.2 and p(z_t+1=1|z_t=2)=1-p(z_t+1=2|z_t=2)=0.1. The equation for p(z3=1|z1=1) can thus be derived as a combination of previous probabilities: [p(z3=1|z2=2)*p(z2=2|z1=1)] + [p(z3=1|z2=1)*p(z2=1|z1=1)]=0.1*0.2+0.8*0.8=0.66. Similarly for p(z3=2|z1=2): [p(z3=2|z2=2)*p(z2=2|z1=2)]+[p(z3=2|z2=1)*p(z2=1|z1=2)]=0.9*0.9+0.2*0.1=0.83. Consequently, the overall probability for using the same die for the initial three rolls can be computed via: {p(z1=1)*p(z3=1|z1=1)}*{p(z1=2)*p(z3=2|z1=2)} = 0.5*0.66+0.5*0.83 = 0.745; thus, the probability amounts to 0.745.
It’s challenging to demonstrate that on here unless I sketch a diagram for you... It’s important to recognize that 10 thousandths equates to 1 hundredth. If you draw a square measuring 100 by 100 centimeters on graph paper, it will include a thousand individual squares because 100 x 100 equals 1000, and if you shade 10 of those squares, that represents 10 thousandths.
Answer:
Step-by-step explanation:
If you compare numbers based on the hundreds digit, the digits to the left (thousands and ten-thousands) remain the same, while the hundreds digit differs.
This applies for any five-digit numbers structured like:
(3)(5)(not 7)(any digit)(any digit).
Digits to the right of the decimal point, as well as those in the tens and ones places, don't impact this comparison.
