SATHEE: Probability 2 Question 14

Probability 2 Question 14

14. $A, B, C$ are events such that

$\begin{matrix} P_{r} (A) = 0.3, P_{r} (B) = 0.4, P_{r} (C) = 0.8 \\ P_{r} (A B) = 0.08, P_{r} (A C) = 0.28 and P_{r} (A B C) = 0.09 \end{matrix}$

If $P_{r} (A \cup B \cup C) \geq 0.75$ , then show that $P_{r} (B C)$ lies in the interval [ $0.23, 0.48$ ].

$(1983, 2 M)$

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Answer:

Correct Answer: 14. (a)

Solution:

We know that,

$\begin{aligned} P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) \\ - P (C \cap A) + P (A \cap B \cap C) = P (A \cup B \cup C) \\ \Rightarrow 0.3 + 0.4 + 0.8 - 0.08 + 0.28 + P (B C) + 0.09 \\ = P (A \cup B \cup C) \end{aligned}$

$\Rightarrow 1.23 - P (B C) = P (A \cup B \cup C)$

where, $0.75 \leq P (A \cup B \cup C) \leq 1$

$\begin{array}{lc} \Rightarrow & 0.75 \leq 1.23 - P (B C) \leq 1 \\ \Rightarrow & - 0.48 \leq - P (B C) \leq - 0.23 \\ \Rightarrow & 0.23 \leq P (B C) \leq 0.48 \end{array}$

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Probability 2 Question 14

14. A,B,C are events such that

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14. $A, B, C$ are events such that