A Numerical Example 2 points possible (graded) Let's say that the prior for the positive class takes the following value: P(y=+) = 0.3 Also, say that P(D|theta+) = .3 and P (D|theta

Question

-) = .6 From the above values of prior and likelihood, calculate the value of P(D), the probability of generating document D. Enter the value below: From P (D) also estimate the posterior probability P(y=+|D). Enter your answer as a numerical expression or round it off to two decimal places.

Answer 1

To calculate \( P(D) \), the probability of generating document \( D \), we can use the law of total probability:

\[ P(D) = P(D | y=+) P(y=+) + P(D | y=-) P(y=-) \]

We are given the following values:

• \( P(y=+) = 0.3 \)
• \( P(y=-) = 1 - P(y=+) = 1 - 0.3 = 0.7 \)
• \( P(D | y=+) = 0.3 \)
• \( P(D | y=-) = 0.6 \)

Now substitute these values into the equation for \( P(D) \):

\[ P(D) = P(D | y=+) P(y=+) + P(D | y=-) P(y=-) \]

\[ P(D) = (0.3)(0.3) + (0.6)(0.7) \]

Calculating each term:

\[ P(D) = 0.09 + 0.42 \]

\[ P(D) = 0.51 \]

Now, to find the posterior probability \( P(y=+ | D) \) using Bayes' theorem:

\[ P(y=+ | D) = \frac{P(D | y=+) P(y=+)}{P(D)} \]

Substituting the values:

\[ P(y=+ | D) = \frac{(0.3)(0.3)}{0.51} \]

Calculating the numerator:

\[ P(y=+ | D) = \frac{0.09}{0.51} \]

Calculating:

\[ P(y=+ | D) \approx 0.1765 \]

Rounding to two decimal places:

\[ P(y=+ | D) \approx 0.18 \]

In summary:

• \( P(D) \approx 0.51 \)
• \( P(y=+ | D) \approx 0.18 \)