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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}

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