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NCERT Exemplar · Q46

Q.If the events AA and BB are independent, then P(A∩B)P(A \cap B) is equal to
(A) P(A)+P(B)P(A) + P(B)
(B) P(A)−P(B)P(A) - P(B)
(C) P(A)⋅P(B)P(A) \cdot P(B)
(D) P(A)P(B)\dfrac{P(A)}{P(B)}

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For independent events, the probability of both occurring is the product of their individual probabilities. So P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B), which corresponds to option (C).

The idea of independence is one of the most intuitive in probability — and also one of the most commonly misunderstood. Two events are independent if knowing that one happened gives you no information about whether the other happened. For example, flipping a coin and rolling a die: the coin landing heads doesn't change the chance of rolling a 4.

Now, how does this translate into a mathematical rule? If events are independent, the chance that both occur should simply be the chance of one times the chance of the other. Why? Because the occurrence of one doesn't "interfere" with the other — there's no adjustment needed.

Let's walk through the reasoning step by step.

  1. Recall the definition of independence. Events AA and BB are independent if and only if

P(A∩B)=P(A)⋅P(B).P(A \cap B) = P(A) \cdot P(B).

This is the formal definition. It captures the idea that the probability of both happening is just the product of their separate probabilities.

  1. Check the options against this definition.

    • Option (A): P(A)+P(B)P(A) + P(B) — this is the formula for the union when events are mutually exclusive, not independent. A common mix-up.
    • Option (B): P(A)−P(B)P(A) - P(B) — this has no standard interpretation in probability for intersection.
    • Option (C): P(A)⋅P(B)P(A) \cdot P(B) — exactly the definition of independence.
    • Option (D): P(A)P(B)\dfrac{P(A)}{P(B)} — this relates to conditional probability P(A∣B)P(A|B) when P(B)≠0P(B) \neq 0, but it's not the intersection.
  2. Why the product makes sense intuitively. …

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