We are given the following information: \[ P(B) = \frac{3}{5}, \quad P(A/B) = \frac{1}{2}, \quad P(A \cup B) = \frac{4}{5}. \] From the conditional probability formula: \[ P(A/B) = \frac{P(A \cap B)}{P(B)}. \] Substituting the given values: \[ \frac{1}{2} = \frac{P(A \cap B)}{\frac{3}{5}}, \] Solving for \( P(A \cap B) \): \[ P(A \cap B) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10}. \] Next, we use the formula for the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \] Substituting the given values: \[ \frac{4}{5} = P(A) + \frac{3}{5} - \frac{3}{10}. \] Multiplying the entire equation by 10 to clear the denominators: \[ 8 = 10P(A) + 6 - 3. \] Simplifying: \[ 8 = 10P(A) + 3 \quad \Rightarrow \quad 10P(A) = 5 \quad \Rightarrow \quad P(A) = \frac{1}{2}. \]
So, the correct answer is (B) : \(\frac{1}{2}\).
Given:
Step 1: Use the conditional probability formula:
\(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
\(\Rightarrow \frac{1}{2} = \frac{P(A \cap B)}{3/5}\)
\(\Rightarrow P(A \cap B) = \frac{1}{2} \cdot \frac{3}{5} = \frac{3}{10}\)
Step 2: Use the formula for union of two events:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(\frac{4}{5} = P(A) + \frac{3}{5} - \frac{3}{10}\)
Simplify:
\(\frac{4}{5} = P(A) + \frac{6}{10} - \frac{3}{10} = P(A) + \frac{3}{10}\)
\(\Rightarrow P(A) = \frac{4}{5} - \frac{3}{10} = \frac{8}{10} - \frac{3}{10} = \frac{5}{10} = \frac{1}{2}\)
Answer: \(\frac{1}{2}\)
A gardener wanted to plant vegetables in his garden. Hence he bought 10 seeds of brinjal plant, 12 seeds of cabbage plant, and 8 seeds of radish plant. The shopkeeper assured him of germination probabilities of brinjal, cabbage, and radish to be 25%, 35%, and 40% respectively. But before he could plant the seeds, they got mixed up in the bag and he had to sow them randomly.
Given three identical bags each containing 10 balls, whose colours are as follows:
Bag I | 3 Red | 2 Blue | 5 Green |
Bag II | 4 Red | 3 Blue | 3 Green |
Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is: