Let's analyze and solve the problem step by step.

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Assumptions and Definitions

1. Lib A = Library A
2. Lib B = Library B
3. Testing Framework = Internal testing framework
4. The probabilities of each event are calculated under the given constraints:
   • Developers are twice as likely to use Lib A compared to Lib B.
   • Developers are twice as likely to use external libraries (A or B) compared to the Testing Framework.
   • A developer cannot use both the Testing Framework and external libraries together.
   • There is a 20% probability that a developer uses both Lib A and Lib B together.

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Step 1: Probability Definitions

Let the probabilities of using Lib A, Lib B, and the Testing Framework be represented as:

• $P(A)$ = Probability of using Lib A.
• $P(B)$ = Probability of using Lib B.
• $P(TF)$ = Probability of using the Testing Framework.

From the question:

1. Lib A is twice as likely as Lib B: $P(A) = 2P(B)$

2. External libraries are twice as likely as the Testing Framework: $P(A \cup B) = 2P(TF)$

3. Probability of using both Lib A and Lib B: $P(A \cap B) = 0.2$

4. The total probability of a developer choosing any option must sum to 1: $P(A \cup B) + P(TF) = 1$

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Step 2: Solve for Probabilities

1. Using $P(A) = 2P(B)$, let $P(B) = x$. Then $P(A) = 2x$.

2. The probability of using external libraries is: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substituting values: $P(A \cup B) = 2x + x - 0.2 = 3x - 0.2$

3. Using the fact that $P(A \cup B) = 2P(TF)$: $3x - 0.2 = 2P(TF)$

4. The total probability is: $P(A \cup B) + P(TF) = 1$ Substituting $P(TF) = \frac{3x - 0.2}{2}$: $3x - 0.2 + \frac{3x - 0.2}{2} = 1$ Solve this equation: $\frac{6x - 0.4 + 3x - 0.2}{2} = 1$ $\frac{9x - 0.6}{2} = 1$ $9x - 0.6 = 2$ $9x = 2.6$ $x = \frac{2.6}{9} = 0.2889$

5. Now we calculate the individual probabilities:

   • $P(B) = x = 0.2889$
   • $P(A) = 2x = 2(0.2889) = 0.5778$
   • $P(TF) = \frac{3x - 0.2}{2} = \frac{3(0.2889) - 0.2}{2} = \frac{0.8667 - 0.2}{2} = 0.3334$.

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Step 3: Solve for Each Question

a. Probability of using Lib B:

$P(B) = 0.2889$

b. Probability of using the Testing Framework:

$P(TF) = 0.3334$

c. Probability of using Lib A given Lib B is in use:

Using the conditional probability formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ Substitute values: $P(A|B) = \frac{0.2}{0.2889} = 0.6923$

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Final Answers

a. $P(B) = 0.2889$
b. $P(TF) = 0.3334$
c. $P(A|B) = 0.6923$

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Follow-Up Questions

1. How would these probabilities change if the chance of using both Lib A and Lib B was different?
2. What is the probability of using neither the Testing Framework nor the external libraries?
3. Can the Testing Framework's usage probability be further broken down into scenarios?
4. How does the assumption that libraries and the Testing Framework cannot be used together simplify the problem?
5. What if $P(A \cup B) \neq 2P(TF)$? How would this affect the results?

Tip: When dealing with probabilities, always check for constraints (e.g., total probability = 1) to validate your results.