Question

A die is thrown once. If the event 'the number obtained on the die is a multiple of 3' is represented by E and 'the number obtained on the die is even' is represented by F, tell whether the events E and F are independent.

Hint:

The concept of independence is crucial in probability. Always remember the test: \(P(E \cap F) = P(E) \times P(F)\). If they are equal, the events are independent. If they are not equal, the events are dependent.

Answer 1

Step 1: Understanding the Concept:
Two events, E and F, are said to be independent if the occurrence of one event does not affect the probability of the other event occurring. Mathematically, two events are independent if and only if \(P(E \cap F) = P(E) \times P(F)\). We need to calculate these three probabilities and check if the condition holds.
Step 2: Key Formula or Approach:
1. Define the sample space S for throwing a single die.
2. Identify the outcomes for event E (a multiple of 3).
3. Identify the outcomes for event F (an even number).
4. Identify the outcomes for the intersection event \(E \cap F\) (an even multiple of 3).
5. Calculate the probabilities \(P(E)\), \(P(F)\), and \(P(E \cap F)\).
6. Check if \(P(E \cap F) = P(E) \times P(F)\).
Step 3: Detailed Explanation:
1. Sample Space:
When a die is thrown once, the sample space is S = \{1, 2, 3, 4, 5, 6\}. The total number of outcomes is \(n(S) = 6\).
2. Event E:
E is the event that the number is a multiple of 3. The outcomes are E = \{3, 6\}.
The number of favorable outcomes is \(n(E) = 2\).
The probability is \(P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}\).
3. Event F:
F is the event that the number is even. The outcomes are F = \{2, 4, 6\}.
The number of favorable outcomes is \(n(F) = 3\).
The probability is \(P(F) = \frac{n(F)}{n(S)} = \frac{3}{6} = \frac{1}{2}\).
4. Intersection Event \(E \cap F\):
\(E \cap F\) is the event that the number is both a multiple of 3 and even. The only outcome is \(E \cap F = \{6\}\).
The number of favorable outcomes is \(n(E \cap F) = 1\).
The probability is \(P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{6}\).
5. Check for Independence:
We need to check if \(P(E \cap F) = P(E) \times P(F)\).
Let's calculate the product of \(P(E)\) and \(P(F)\): \[ P(E) \times P(F) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \] We see that \(P(E \cap F) = \frac{1}{6}\), which is equal to \(P(E) \times P(F)\).
Step 4: Final Answer:
Since \(P(E \cap F) = P(E) \times P(F)\), the events E and F are independent.