Let E and F are events of a experiment such that P(E) = 3/10 P(F) = ½ and P(F|E) = ⅖. Find the value of (i) P(E ∩ F) (ii) P(E|F) (iii) P(E U F)

1. Break Down the Problem

We need to find the following probabilities based on the provided information:

1. $P(E \cap F)$ - the probability of both events E and F occurring.
2. $P(E | F)$ - the conditional probability of E given F.
3. $P(E \cup F)$ - the probability of either event E or F occurring.

2. Relevant Concepts

We will use the following probability concepts and formulas:

• The formula for the probability of the intersection of two events: $P(E \cap F) = P(F | E) \cdot P(E)$
• The formula for conditional probability: $P(E | F) = \frac{P(E \cap F)}{P(F)}$
• The formula for the probability of the union of two events: $P(E \cup F) = P(E) + P(F) - P(E \cap F)$

3. Analysis and Detail

(i) Calculate $P(E \cap F)$

Given:

• $P(E) = \frac{3}{10}$
• $P(F | E) = \frac{2}{5}$

Using the intersection formula: $P(E \cap F) = P(F | E) \cdot P(E) = \left(\frac{2}{5}\right) \cdot \left(\frac{3}{10}\right) = \frac{6}{50} = \frac{3}{25}$

(ii) Calculate $P(E | F)$

Using the conditional probability formula: $P(E | F) = \frac{P(E \cap F)}{P(F)}$ Given:

• $P(F) = \frac{1}{2}$

We already found $P(E \cap F) = \frac{3}{25}$: $P(E | F) = \frac{\frac{3}{25}}{\frac{1}{2}} = \frac{3}{25} \cdot \frac{2}{1} = \frac{6}{25}$

(iii) Calculate $P(E \cup F)$

Using the union formula: $P(E \cup F) = P(E) + P(F) - P(E \cap F)$ Substituting the known values: $P(E) = \frac{3}{10}, \quad P(F) = \frac{1}{2}, \quad P(E \cap F) = \frac{3}{25}$

Calculating $P(E) + P(F)$: $P(E) + P(F) = \frac{3}{10} + \frac{1}{2} = \frac{3}{10} + \frac{5}{10} = \frac{8}{10} = \frac{4}{5}$

Now subtracting $P(E \cap F)$: $P(E \cup F) = \frac{4}{5} - \frac{3}{25}$ To perform this subtraction, let's get a common denominator: $\frac{4}{5} = \frac{20}{25}$ Thus, $P(E \cup F) = \frac{20}{25} - \frac{3}{25} = \frac{17}{25}$

4. Verify and Summarize

All calculations have been checked and are consistent with the formulas used.

Final Answer

1. $P(E \cap F) = \frac{3}{25}$
2. $P(E | F) = \frac{6}{25}$
3. $P(E \cup F) = \frac{17}{25}$