/sci/ - Science & Math

Congrats /sci/, my first post in over a year

Suppose we have [math] M [/math] boxes and [math] N [/math] boxes contain golden balls. Label these [math] N_k [/math] and define [math] G(N_k) [/math] and [math] T(N_k) [/math] as the golden and total balls in [math] N_k [/math] respectively.

Note that
[eqn] \text{Probability of drawing a ball from} \ N_k = P(N_k) = \frac{1}{N}\frac{G(N_k)}{T(N_k)}[/eqn]

Now suppose we drew a golden ball. The probability it came from [math] N_k [/math] is the PERCENTAGE of the probability [math] N_k [/math] accounts for in the total possibility of drawing a golden ball. In other words
[eqn] \text{Probability ball came from} \ N_k = Q(N_k) = \frac{P(N_k)}{P(N_1)+P(N_2)+P(N_3)\cdots}[/eqn]

Finally the probability of drawing another golden ball would be

[eqn] \sum_k Q(N_k)\frac{G(N_k)-1}{T(N_k)-1} [/eqn]
as this is the summation of the probability we picked boxed [math] N_k [/math] multiplied by the probability of drawing another golden ball.

Congratulations, you just learned the basics of Bayesian probability.

Congrats /sci/, my first post in over a year

Suppose we have [math] M [/math] boxes and [math] N [/math] boxes contain golden balls. Label these [math] N_k [/math] and define [math] G(N_k) [/math] and [math] T(N_k) [/math] as the golden and total balls in [math] N_k [/math] respectively.

Note that
[eqn] \text{Probability of drawing a ball from} \ N_k = P(N_k) = \frac{1}{N}\frac{G(N_k)}{T(N_k)}[/eqn]

Now suppose we drew a golden ball. The probability it came from [math] N_k [/math] is the PERCENTAGE of the probability [math] N_k [/math] accounts for in the total possibility of drawing a golden ball. In other words
[eqn] \text{Probability ball came from} \ N_k = Q(N_k) = \frac{P(N_k)}{P(N_1)+P(N_2)+P(N_3)\cdots}[/eqn]

Finally the probability of drawing another golden ball would be

[eqn] \sum_k Q(N_k)\frac{G(N_k)-1}{T(N_k)-1} [/eqn]
as this is the summation of the probability we picked boxed [math] N_k [/math] multiplied by the probability of drawing another golden ball.

Congratulations, you just learned the basics of Bayesian probability.

Congrats /sci/, my first post in over a year

Suppose we have [math] M [/math] boxes and [math] N [/math] boxes contain golden balls. Label these [math] N_k [/math] and define [math] G(N_k) [/math] and [math] T(N_k) [/math] as the golden and total balls in [math] N_k [/math] respectively.

Note that
[eqn] \text{Probability of drawing a ball from} \ N_k = P(N_k) = \frac{1}{N}\frac{G(N_k)}{T(N_k)}[/eqn]

Now suppose we drew a golden ball. The probability it came from [math] N_k [/math] is the PERCENTAGE of the probability [math] N_k [/math] accounts for in the total possibility of drawing a golden ball. In other words
[eqn] \text{Probability ball came from} \ N_k = Q(N_k) = \frac{P(N_k)}{P(N_1)+P(N_2)+P(N_3)\cdots}[/eqn]

Finally the probability of drawing another golden ball would be

[eqn] \sum_k Q(N_k)\frac{G(N_k)-1}{T(N_k)-1} [/eqn]
as this is the summation of the probability we picked boxed [math] N_k [/math] multiplied by the probability of drawing another golden ball.

Congratulations, you just learned the basics of Bayesian probability.