A, B, C are events such that P(A) =...

MathematicsAddition and Multiplication TheoremsJEE Advanced 1983Easy

$A, B, C$ are events such that $P (A) = 0.3, P (B) = 0.4, P (C) = 0.8, P (A B) = 0.08, P (A C) = 0.28, P (A B C) = 0.09$ . If $P (A \cup B \cup C) \geq 0.75$ , then show that $P (B C)$ lies in the interval $0.23 \leq x \leq 0.48$ .

Visualized Solution (English)

Visualizing the Events .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A, B, C

Given events $A, B, C$ with their respective probabilities.
$P (A) = 0.3, P (B) = 0.4, P (C) = 0.8$
$P (A B) = 0.08, P (A C) = 0.28, P (A B C) = 0.09$
We need to find the interval for $P (B C)$ .

The Inclusion-Exclusion Principle

Apply the Inclusion-Exclusion Principle for three events:
$P (A \cup B \cup C) = P (A) + P (B) + P (C) - [P (A B) + P (B C) + P (C A)] + P (A B C)$

Substituting Known Values

Substitute the given values into the formula:
$P (A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.08 - P (B C) - 0.28 + 0.09$

Simplifying the Expression

Summing the constants:
$0.3 + 0.4 + 0.8 + 0.09 = 1.59$
Subtracting the known intersections:
$1.59 - 0.08 - 0.28 = 1.23$
Simplified Equation:
$P (A \cup B \cup C) = 1.23 - P (B C)$

Applying Probability Bounds

We know the fundamental bounds of probability:
$P (A \cup B \cup C) \leq 1$
And the given condition:
$P (A \cup B \cup C) \geq 0.75$
Combining these, we get:
$0.75 \leq 1.23 - P (B C) \leq 1$

Solving the Inequality

Subtract $1.23$ from all sides:
$0.75 - 1.23 \leq - P (B C) \leq 1 - 1.23$
$- 0.48 \leq - P (B C) \leq - 0.23$

Final Range for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (B C)

Multiply by $- 1$ and reverse the inequality signs:
$0.48 \geq P (B C) \geq 0.23$
Rearranging for the final interval:
$0.23 \leq P (B C) \leq 0.48$
Hence Proved.

Key Takeaways

Key Takeaway: Always remember the implicit bound $P (Union) \leq 1$ .
Next Challenge: What happens to the range of $P (B C)$ if $P (A \cup B \cup C)$ was exactly $0.8$ ?

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