If A and B are two independent events, then show that the probabilit

1.	If A and B are two independent events, then show that the probability of occurrence of at least one of and is given by 1 − P(A′)P(B′).
Answer» Given that events and are two independent events. Therefore, P(A ∩ B) = P(A)P(B). ...(1) Now, the probability of occurrence of at least one of A and B is P(A ∪ B). We know that P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A)P(B)(By equation 1) ⇒ P(A ∪ B) = P(A) + P(B)[1 − P(A)] = P(A) + P(B)P(A′) ( \(\because\) P(A′) = 1 − P(A)) ⇒ P(A ∪ B) = 1 − 1 + P(A) + P(B)P(A′) = 1 − [1 − P(A)] + P(B)P(A′) ⇒ P(A ∪ B) = 1 − P(A ′ ) + P(B)P(A ′ ) = 1 − P(A′)[1 − P(B)] ( \(\because\) P(A ′) = 1 − P(A)) ⇒ P(A ∪ B) = 1 − P(A ′ )P(B ′ ). Hence, if A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1 - P(A′) P(B′ ). (Hence prove)

If A and B are two independent events, then show that the probability of occurrence of at least one of and is given by 1 − P(A′)P(B′).

Answer»

Given that events and are two independent events.

Therefore, P(A ∩ B) = P(A)P(B). ...(1)

Now, the probability of occurrence of at least one of A and B is P(A ∪ B).

We know that P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A)P(B)(By equation 1)

⇒ P(A ∪ B) = P(A) + P(B)[1 − P(A)] = P(A) + P(B)P(A′) ( \(\because\) P(A′) = 1 − P(A))

⇒ P(A ∪ B) = 1 − 1 + P(A) + P(B)P(A′) = 1 − [1 − P(A)] + P(B)P(A′)

⇒ P(A ∪ B) = 1 − P(A ′ ) + P(B)P(A ′ ) = 1 − P(A′)[1 − P(B)] ( \(\because\) P(A ′) = 1 − P(A))

⇒ P(A ∪ B) = 1 − P(A ′ )P(B ′ ).

Hence, if A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1 - P(A′) P(B′ ).

(Hence prove)

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