A, B , and C are three events such that P(A)=0.6 , P(B)=0.4, P(C)=0.5, P(A B)=0.8, P(A C)=0.3 and P(A B C)=0.2 and P(A B C)≥ 0.85 . What is the maximum value of P(B C) ?

Correct answer: C. 0.35 aligned P( A)=0.6, P( B) & =0.4, P(C)=0.5 P( A B) & =P( A)+P( B)-P( A B) & =0.6+0.4-0.8=0.2 aligned Now, aligned P( A B C)= & P( A)+P( B)+P(C) & -P( A B)-P( A C) & -P( B C)+P( A B C) = & 1.5-0.2-0.3-P( B C)+0.2 aligned array rlrl & P(A B C) & =1.2-P(B C) & 0.85 & ≤ P(A B C) ≤ 1 & 0.85 & ≤ 1.2-P(B C) ≤ 1 & -0.35 & ≤-P(B C) ≤-0.2 & 0.2 & ≤ P(B C) ≤ 0.35 array Maximum value of P(B C)=0.35

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Probability

NDA 2024 · Mathematics · Probability

A, B , and C are three events such that P(A)=0.6 , P(B)=0.4,...

\mathrm{A}, \mathrm{B}

, and C are three events such that

\mathrm{P}(\mathrm{A})=0.6

\mathrm{P}(\mathrm{B})=0.4, \mathrm{P}(\mathrm{C})=0.5, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.8, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=0.3

and

\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0.2

and

\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\geq 0.85

. What is the maximum value of

\mathrm{P}(\mathrm{B} \cap \mathrm{C})

Options

A0.1
B0.2
C0.35
D0.45

Correct answer

C. 0.35

Step-by-step solution

\begin{aligned} \mathrm{P}(\mathrm{ A})=0.6, \mathrm{P}(\mathrm{ B}) & =0.4, \mathrm{P}(\mathrm{C})=0.5 \\ \mathrm{P}(\mathrm{ A} \cap \mathrm{ B}) & =\mathrm{P}(\mathrm{ A})+\mathrm{P}(\mathrm{ B})-\mathrm{P}(\mathrm{ A} \cup \mathrm{ B}) \\ & =0.6+0.4-0.8=0.2 \end{aligned}

Now,

\begin{aligned} \mathrm{P}(\mathrm{ A} \cup \mathrm{ B} \cup \mathrm{C})= & \mathrm{P}(\mathrm{ A})+\mathrm{P}(\mathrm{ B})+\mathrm{P}(\mathrm{C}) \\ & -\mathrm{P}(\mathrm{ A} \cup \mathrm{ B})-\mathrm{P}(\mathrm{ A} \cap \mathrm{C}) \\ & -\mathrm{P}(\mathrm{ B} \cap \mathrm{C})+\mathrm{P}(\mathrm{ A} \cap \mathrm{ B} \cap \mathrm{C}) \\ = & 1.5-0.2-0.3-\mathrm{P}(\mathrm{ B} \cap \mathrm{C})+0.2 \end{aligned}

\begin{array}{rlrl}\Rightarrow & \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) & =1.2-\mathrm{P}(\mathrm{B} \cap \mathrm{C}) \\ \because & 0.85 & \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \leq 1 \\ \Rightarrow & 0.85 & \leq 1.2-\mathrm{P}(\mathrm{B} \cap \mathrm{C}) \leq 1 \\ \Rightarrow & -0.35 & \leq-\mathrm{P}(\mathrm{B} \cap \mathrm{C}) \leq-0.2 \\ \Rightarrow & 0.2 & \leq \mathrm{P}(\mathrm{B} \cap \mathrm{C}) \leq 0.35\end{array}

Maximum value

of \mathrm{P}(\mathrm{B} \cap \mathrm{C})=0.35

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