Given the probabilities for events $A$ and $B$, we need to find $P(A \mid B)$, which represents the conditional probability of $A$ given $B$.

Given:

1. $P(A \cap B') = \frac{1}{5}$
2. $P(A' \cap B) = \frac{3}{20}$
3. $P(A' \cap B') = \frac{1}{10}$

To Find:

• $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

Step 1: Calculate $P(B')$

Using the total probability: $P(B') = P(A \cap B') + P(A' \cap B')$ $P(B') = \frac{1}{5} + \frac{1}{10} = \frac{2}{10} + \frac{1}{10} = \frac{3}{10}$

Step 2: Calculate $P(B)$

Since $P(B') = 1 - P(B)$, we have: $P(B) = 1 - P(B') = 1 - \frac{3}{10} = \frac{7}{10}$

Step 3: Use Total Probability to find $P(A)$

The total probability $P(A)$ can be found as: $P(A) = P(A \cap B) + P(A \cap B')$ We already know $P(A \cap B') = \frac{1}{5}$, and we need to express $P(A \cap B)$.

Step 4: Find $P(A)$ using the complement

$P(A') = P(A' \cap B) + P(A' \cap B')$ $P(A') = \frac{3}{20} + \frac{1}{10} = \frac{3}{20} + \frac{2}{20} = \frac{5}{20} = \frac{1}{4}$ So: $P(A) = 1 - P(A') = 1 - \frac{1}{4} = \frac{3}{4}$

Step 5: Calculate $P(A \cap B)$

Now, use the total probability for $A$: $P(A) = P(A \cap B) + P(A \cap B')$ $\frac{3}{4} = P(A \cap B) + \frac{1}{5}$ $P(A \cap B) = \frac{3}{4} - \frac{1}{5} = \frac{15}{20} - \frac{4}{20} = \frac{11}{20}$

Step 6: Find $P(A \mid B)$

Now, find $P(A \mid B)$: $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{11}{20}}{\frac{7}{10}} = \frac{11}{20} \times \frac{10}{7} = \frac{11}{14}$

So, $P(A \mid B) = \frac{11}{14}$.

Would you like more details or have any other questions?

Related Questions:

1. How can you derive $P(A \mid B)$ from given joint and marginal probabilities?
2. What is the significance of conditional probability in decision making?
3. How can we verify the consistency of given probability values?
4. What is the relationship between $P(A \mid B)$ and $P(B \mid A)$?
5. How do independent events affect the calculation of conditional probabilities?

Tip: When calculating conditional probabilities, make sure the probabilities used for the condition (like $P(B)$) are accurate and account for all possible scenarios.