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Exercise 13.2 · Q6

Q.Let E and F be events with P(E)=35P(E) = \frac{3}{5}, P(F)=310P(F) = \frac{3}{10} and P(E∩F)=15P(E \cap F) = \frac{1}{5}. Are E and F independent?

Rajasthan RbseTextbookSubjective· 3mImportance★★★★★
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Events E and F are independent if P(E∩F)=P(E)⋅P(F)P(E \cap F) = P(E) \cdot P(F). Here P(E∩F)=15P(E \cap F) = \frac{1}{5} and P(E)⋅P(F)=950P(E) \cdot P(F) = \frac{9}{50}, which are not equal, so E and F are not independent.

The idea of independence is simple: two events are independent if knowing that one has occurred gives you no information about whether the other has occurred. In probability terms, this means the probability that both happen together is exactly the product of their individual probabilities. That’s the test we’ll use.

Let’s check it step by step.

  1. Write down what’s given. We have

P(E)=35,P(F)=310,P(E∩F)=15.P(E) = \frac{3}{5}, \quad P(F) = \frac{3}{10}, \quad P(E \cap F) = \frac{1}{5}.

  1. Compute the product P(E)⋅P(F)P(E) \cdot P(F). Multiply the two probabilities:

P(E)⋅P(F)=35×310=950.P(E) \cdot P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50}.

  1. Compare with P(E∩F)P(E \cap F). The given intersection probability is 15\frac{1}{5}. Write it with denominator 50 to compare easily:

15=1050.\frac{1}{5} = \frac{10}{50}.

So we have P(E∩F)=1050P(E \cap F) = \frac{10}{50} and P(E)⋅P(F)=950P(E) \cdot P(F) = \frac{9}{50}.

  1. Apply the independence condition. For independence, we need P(E∩F)=P(E)⋅P(F)P(E \cap F) = P(E) \cdot P(F). Here 1050≠950\frac{10}{50} \neq \frac{9}{50}, so the condition fails. …

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