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  • Need solution for RD Sharma Maths Class12 Chapter 30 Probability Exercise Fill in the Blanks Question, Question 16.

Need solution for RD Sharma Maths Class12 Chapter 30 Probability Exercise Fill in the Blanks Question, Question 16.

Answers (1)

Answer : \frac{ x+y+z}{2}

Given:

Let A,B,C bo there events such that  P(A\cap B\cap C)=0
P(Exactly one of A and B occurs) =x,

P (Exactly one of B and C occurs) =y
P (Exactly one of A and C occurs )=z.

Then P(A\cup B\cup C)= ................

Hint:

Using the formula of P(A\cup B\cup C)

Explanation:

\begin{aligned} &\mathrm{P}(\mathrm{A} \text { or } \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{x} \\ &\mathrm{P}(\mathrm{B} \text { or } \mathrm{C})=\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{y} \\ &\mathrm{P}(\mathrm{A} \text { or } \mathrm{C})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{C})=2 \end{aligned}

Adding these three we get,

\begin{aligned} &2 \mathrm{P}(\mathrm{A})+2 \mathrm{P}(\mathrm{B})+2 \mathrm{P}(\mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})-2 \mathrm{P}(\mathrm{B} \cap \mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\mathrm{x}+\mathrm{y}+\mathrm{z} \\ &\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{B} \cap \mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{2}-(1) \end{aligned}

We know,
\begin{aligned} &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) \\ &\Rightarrow P(A \cup B \cup C)=\frac{x+y+2}{2}+P(A \cap B \cap C) \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \text { of }(1)] \end{aligned}

But  P(A\cap B\cap C)=0
\therefore P(A \cup B \cup C)=\frac{x+y+2}{2}+0=\frac{x+y+z}{2}

Posted by

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