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Step-by-Step Solution
Step 1: List the given probabilities
We are given that for three events A, B, and C (with 0 < k < 1):
The probability that exactly one of A and B occurs is (1 − k).
The probability that exactly one of B and C occurs is (1 − 2k).
The probability that exactly one of C and A occurs is (1 − k).
The probability that all A, B, and C occur simultaneously is $k^2$.
We want to find the probability that at least one of A, B, or C occurs, i.e., $P(A \cup B \cup C)$.
Step 2: Express the “exactly one event occurs” probabilities in terms of pairwise intersections
Recall that the probability of exactly one of A and B happening is:
$P(\overline{A} \cap B) + P(A \cap \overline{B}) = P(A) + P(B) - 2\,P(A \cap B).$
From the problem statement:
$P(A) + P(B) - 2\,P(A \cap B) = 1 - k \quad \dots \text{(i)}$
$P(B) + P(C) - 2\,P(B \cap C) = 1 - k \quad \dots \text{(ii)}$
$P(C) + P(A) - 2\,P(C \cap A) = 1 - 2k \quad \dots \text{(iii)}$
Step 3: Sum the equations to relate $P(A \cap B \cap C)$
Adding (i), (ii), and (iii):
$\bigl[P(A) + P(B) - 2\,P(A \cap B)\bigr] + \bigl[P(B) + P(C) - 2\,P(B \cap C)\bigr] + \bigl[P(C) + P(A) - 2\,P(C \cap A)\bigr] = (1 - k) + (1 - k) + (1 - 2k).$
Simplifying the left-hand side gives:
$2\,\bigl[P(A) + P(B) + P(C)\bigr] - 2\,\bigl[P(A \cap B) + P(B \cap C) + P(C \cap A)\bigr].
The right-hand side totals $1 - k + 1 - k + 1 - 2k = 3 - 4k.$
Hence,
$2\,[P(A) + P(B) + P(C)] - 2\,[P(A \cap B) + P(B \cap C) + P(C \cap A)] = 3 - 4k.$
Divide both sides by 2:
$P(A) + P(B) + P(C) - \bigl[P(A \cap B) + P(B \cap C) + P(C \cap A)\bigr] = \frac{3 - 4k}{2} \quad \dots \text{(iv)}$
Step 4: Use the inclusion-exclusion formula for three events
The probability of at least one of A, B, or C occurring is:
$P(A \cup B \cup C) = P(A) + P(B) + P(C)\;-\;P(A \cap B) - P(B \cap C) - P(C \cap A)\;+\;P(A \cap B \cap C).$
From (iv), the term $P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)]$ equals $\frac{3 - 4k}{2}.$ Given $P(A \cap B \cap C) = k^2,$ we get:
$P(A \cup B \cup C) = \frac{3 - 4k}{2} + k^2.$
Step 5: Show that $P(A \cup B \cup C)$ is > 1/2
Rewrite $P(A \cup B \cup C)$:
$P(A \cup B \cup C) = \frac{3 - 4k + 2k^2}{2} = \frac{2(k - 1)^2 + 1}{2}.$
Notice that $(k - 1)^2 \geq 0$ for all k, so $2(k - 1)^2 + 1 \geq 1.$ Indeed, for $0 < k < 1,$ $(k - 1)^2$ is positive, thus the numerator exceeds 1, making:
$P(A \cup B \cup C) > \frac{1}{2}.$
Hence, the probability that at least one of A, B, and C occurs is greater than 1/2.
