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Q6:
Let E and F be events with $P(E) = \frac{3}{5}$, $P(F) = \frac{3}{10}$ and $P(E \cap F) = \frac{1}{5}$. Are E and F independent?

Solution :

Step 1: Establishing the Theoretical Foundation for Independence

Two events $E$ and $F$ associated with the same sample space are defined as independent if and only if the probability of their intersection is equal to the product of their individual probabilities. Mathematically, the condition for independence is defined as follows:

$P(E \cap F) = P(E) \times P(F)$

[Reference: Definition of Independent Events in Probability Theory]

Step 2: Identifying Given Variables

From the problem statement, we have the following assigned values:

  • $P(E) = \frac{3}{5}$
  • $P(F) = \frac{3}{10}$
  • $P(E \cap F) = \frac{1}{5}$

Step 3: Calculating the Product of Individual Probabilities

To evaluate independence, we compute the product $P(E) \times P(F)$ using the values identified in Step 2:

$P(E) \times P(F) = \left( \frac{3}{5} \right) \times \left( \frac{3}{10} \right)$

Performing the multiplication of numerators and denominators:

$P(E) \times P(F) = \frac{3 \times 3}{5 \times 10} = \frac{9}{50}$

Step 4: Comparative Analysis

We compare the calculated product with the given value of $P(E \cap F)$:

  • $P(E \cap F) = \frac{1}{5} = \frac{10}{50}$
  • $P(E) \times P(F) = \frac{9}{50}$

Since $\frac{10}{50} \neq \frac{9}{50}$, it follows that $P(E \cap F) \neq P(E) \times P(F)$.

Visual Representation of Probability Sets

E F E∩F Sample Space S

Conclusion

Because the condition $P(E \cap F) = P(E) \times P(F)$ is not satisfied, the events $E$ and $F$ do not meet the criteria for independence.

Final Solution: No, E and F are not independent because $P(E \cap F) \neq P(E) \times P(F)$.


More Questions from Class 12 Mathematics Probability EXERCISE 13.2


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