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: \(\frac{1}{2}\)
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:
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.
In a practical examination, the following pedigree chart was given as a spotter for identification. The students identify the given pedigree chart as 