Question

1. imagine youre managing a software development project. you want to estimate the probability of successfully completing the project within a given time frame, based on past experiences. you have observed the following:

• in similar projects, 70% of the time, they were completed successfully within the deadline.
• youve also tracked the outcomes and found that in 8 out of 10 similar projects, success was achieved.

suppose want to know the probability that the project will succeed with a probability greater than 0.7.

a) find \\(\alpha\\) and \\(\beta\\).

b) set up the pdf needed to find this probability and then set up the integral.

c) use r or some other technology to determine the probability.

Explanation:

Identify prior distribution parameters

We model the success probability \(p\) using a Beta prior distribution, \(\text{Beta}(\alpha_0, \beta_0)\).
The prior mean is given as \(0.70\):

$$\frac{\alpha_0}{\alpha_0 + \beta_0} = 0.7$$

Assuming a base prior representing a single project trial equivalent where \(\alpha_0 + \beta_0 = 1\) (or using the standard default prior setup for updating), we find:

$$\alpha_0 = 0.7, \quad \beta_0 = 0.3$$

Update parameters with observed data

We observe \(x = 8\) successes out of \(n = 10\) trials.
Using the conjugate Beta-Binomial update rules:

$$\alpha = \alpha_0 + x = 0.7 + 8 = 8.7$$

$$\beta = \beta_0 + (n - x) = 0.3 + 2 = 2.3$$

Formulate the posterior PDF

The posterior distribution of the success probability \(p\) is \(\text{Beta}(8.7, 2.3)\).
The probability density function (pdf) is:

$$f(p) = \frac{1}{\text{B}(8.7, 2.3)} p^{7.7} (1-p)^{1.3}, \quad 0 \le p \le 1$$

where \(\text{B}(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\).

Set up the probability integral

We want to find the probability that the project succeeds with a probability greater than \(0.7\):

$$P(p > 0.7) = \int_{0.7}^{1} \frac{1}{\text{B}(8.7, 2.3)} p^{7.7} (1-p)^{1.3} \, dp$$

Calculate the probability using technology

Using R or standard numerical integration for the Beta distribution:
1 - pbeta(0.7, 8.7, 2.3)
Evaluating this yields:

$$P(p > 0.7) \approx 0.7957$$

Answer:

Question a

\(\alpha = 8.7\) and \(\beta = 2.3\)

Question b

The probability density function (pdf) is:

$$f(p) = \frac{\Gamma(11)}{\Gamma(8.7)\Gamma(2.3)} p^{7.7} (1-p)^{1.3}, \quad 0 \le p \le 1$$

The integral to find the probability is:

$$P(p > 0.7) = \int_{0.7}^{1} \frac{\Gamma(11)}{\Gamma(8.7)\Gamma(2.3)} p^{7.7} (1-p)^{1.3} \, dp$$

Question c

Using R:
1 - pbeta(0.7, 8.7, 2.3)
The probability is approximately \(0.7957\) (or \(79.57\%\)).