If P (A) = a and P (B) = b, then show that P (A | B) ≥ (a + b − 1)/b
Question
If P (A) = a and P (B) = b, then show that P (A | B) ≥ (a + b − 1)/b
Solution
Sure, I can help with that. Here's how you can show that P(A | B) ≥ (a + b - 1)/b:
The conditional probability P(A | B) is defined as P(A ∩ B) / P(B).
We know that P(A ∩ B) = P(A) + P(B) - P(A ∪ B).
Since P(A ∪ B) ≤ 1, we have P(A ∩ B) ≥ P(A) + P(B) - 1.
Substituting this into the definition of conditional probability gives:
P(A | B) = P(A ∩ B) / P(B) ≥ (P(A) + P(B) - 1) / P(B) = (a + b - 1) / b.
So, we have shown that P(A | B) ≥ (a + b - 1)/b.
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