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TykiMikk · Answer 1 · 2024-12-24T20:05:12+0000

The probability
$( P(A \cap B \cap C) ) is ( (1)/(24) ).$

The probability of the intersection of three independent events A, B, and C is given by the product of their individual probabilities. Therefore, the probability ( P(A \cap B \cap C) ) is calculated as:

$[ P(A \cap B \cap C) = P(A) * P(B) * P(C) ]$

Substituting the given probabilities, we have:

$[ P(A \cap B \cap C) = (1)/(3) * (1)/(4) * (1)/(2) ]\\[ P(A \cap B \cap C) = (1)/(24) ]$

Therefore, the probability
$( P(A \cap B \cap C) ) is ( (1)/(24) ).$

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