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  • Prove that (i) P(A)= P(AՈB) + P(AՈ ) (ii) P(AՍB) = P(AՈB) + P(AՈ ) + P( ՈB)

Prove that
(i) P(A)= P(A\cap B) + P(A\cap \bar{B})
(ii) P(A\cup B) = P(A\cap B) + P(A\cap \bar{B} ) + P( \bar{A}\cap B)

Answers (1)

It has to be proven that P(A)=P(A \cap B)+P\left(A \cap {\bar{B}}\right)$
As we know,  A= A \cap S
Therefore,
\\\mathrm{s}=\mathrm{A} \cup \mathrm{A}^{\prime}$ and \\$\mathrm{s}=\mathrm{B} \cup \mathrm{B'}$
\\A=A \cap\left(B \cup B^{\prime}\right)$ \\$=(A \cap B) \cup\left(A \cap B^{\prime}\right)$

When two events cannot occur at the same time, they are called mutually exclusive or disjoint events. 

Since (A \cap B) means A and B both occurring at the same time and  

(A \cap B \textsuperscript{'}) means A and B\textsuperscript{'} both occurring at the same time. 

Therefore, it is not possible that (A \cap B) and (A \cap B\textsuperscript{'}) occur at the same time. 

Hence, (A \cap B) and (A \cap B\textsuperscript{'}) are mutually exclusive. 

∴ P[(A \cap B) \cap (A \cap B’)] = 0 ….. (1)

Therefore, A = A \cap (B \cup B’)

Knowing that P (A \cup B) = P (A) + P(B) - P (A \cap B)

\begin{aligned} &\mathrm{P}(\mathrm{A})=\mathrm{P}\left[(\mathrm{A} \cap \mathrm{B}) \cup\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)\right]\\ &=P(A \cap B)+P\left(A \cap B^{\prime}\right)-P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right)\right]\\ &\text { From }(1)\\ &P(A)=P(A \cap B)+P(A \cap \bar{B}) \end{aligned}

Hence proved.

ii) It is to be proven that, P(A \cup B)=P(A \cap B)+P(A \cap \bar{B})+P(\bar{A} \cap B)

A\cupB means the all the possible outcomes of both A and B.
From the above Venn diagram,

\mathrm{AUB}=(\mathrm{A} \cap \mathrm{B}) \cup(\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})

When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.

Since (A \cap B) means A and B both occurring at the same time and

(A \cap B’) means A and B’ both occurring at the same time.

(A' \cap B) means A’ and B both occurring at the same time.

Therefore, it is not possible that (A \cap B), (A \cap B’) and (\bar{A} B) occur at the same time.
Hence these events are mutually exclusive.

\\ P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right)\right]=0 \ldots . .(1)$ \\$P\left[\left(A \cap B^{\prime}\right) \cap P(\bar{A} \cap B)\right]=0 \ldots . .(2)$ \\$P[(A \cap B) \cap P(\bar{A} \cap B)]=0 \ldots . .(3)$ \\$P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right) \cap P(\bar{A} \cap B)\right]=0 \ldots .(4)$ \\$P(A \cup B)=P[(A \cap B) \cup(A \cap \bar{B}) \cup(\bar{A} \cap B)]$
It is known that,
\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} \cap \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})$
Therefore,

\\P(A U B)=P[(A \cap B) \cup(A \cap \bar{B}) \cup(\bar{A} \cap B)]=P(A \cap B)+P(A \cap \bar{B})+P(\bar{A} \cap B)-P[(A \cap B) \cap(A \cap \bar{B})]-$$P[(A \cap B) \cap(\bar{A} \cap B)]-P[(A \cap \bar{B}) \cap(\bar{A} \cap B)]+P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right) \cap P(\bar{A} \cap B)\right]$
From (1),(2),(3) and (4) we get,

P(A U B)=P(A \cap B)+P\left(A \cap {\bar{B}}\right)+P\left(A \cap{ B}\right)$

Hence Proved

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