Math, asked by makhzansaleem, 8 months ago

Let A and B be any two events defined on a sample space S with P(AUB)=3/4, P(A bar) =2/3 and P(A intersection B) =1/4, find P(A) ,P(B) and P(A intersection B bar)​

Answers

Answered by BrainlyPopularman
45

GIVEN :

P(AUB) = ¾

  \:  \:  \bf P({ \overline{A}}) =  \dfrac{2}{3}

  \:  \:  \bf P( \:A \cap B) =  \dfrac{1}{4}

TO FIND :

P(A) , P(B) &   \:  \:  \bf P( \: { \overline{ A \cap B}} \: ) = ?

SOLUTION :

• We know that –

  \implies  \bf P(A) = 1 -  P({ \overline{A}})

• So that –

  \implies  \bf P(A) = 1 -  \dfrac{2}{3}

  \implies  \bf P(A) =  \dfrac{3 - 2}{3}

  \implies \large { \boxed{ \bf P(A) =  \dfrac{1}{3}}}

 \rule{220}{2}

• Now Let's fine   \:  \:  \bf P( \: { \overline{ A \cap B}} \: )

  \implies \bf P( \: { \overline{ A \cap B}} \: ) = 1 - P( \:A \cap B)

  \implies \bf P( \: { \overline{ A \cap B}} \: ) = 1 -  \dfrac{1}{4}

  \implies \bf P( \: { \overline{ A \cap B}} \: ) = \dfrac{4 - 1}{4}

  \implies \large{ \boxed{ \bf P( \: { \overline{ A \cap B}} \: ) = \dfrac{3}{4}}}

 \rule{220}{2}

▪︎ Now Let's find P(B) –

  \implies \large{ \boxed{ \bf P(A \cup B) = P(A) + P(B) -  P( \:A \cap B)}}

• Put the values –

  \implies  \bf  \dfrac{3}{4}  =  \dfrac{1}{3}  + P(B) -   \dfrac{1}{4}

  \implies  \bf  \dfrac{3}{4} +\dfrac{1}{4}  =  \dfrac{1}{3}  + P(B)

  \implies  \bf 1=  \dfrac{1}{3}  + P(B)

  \implies  \bf P(B)= 1 -  \dfrac{1}{3}

  \implies  \bf P(B)=\dfrac{3 - 1}{3}

  \implies  \large{ \boxed{ \bf P(B)=\dfrac{2}{3}}}

 \rule{220}{2}

Answered by Mounikamaddula
24

Answer:

Given:

p(AUB)=3/4,

P(Ā)=2/3

p(AnB)=1/4

To find:

P(A), P(B) and P(ĀnBbar)

Solution:

As we know that,

1)P(A)+P(Ā)=1

P(A)+2/3=1

P(A)=1-2/3

P(A)=1/3

2) From the addition theorem,

P(AUB)=P(A)+P(B)-P(AnB)

3/4=1/3+p(B)-1/4

P(B)=3/4+1/4-1/3

P(B)=1-1/3

P(B)=2/3

3)P(ĀnB bar)=1-P(AUB)=1-3/4=1/4

Step-by-step explanation:

Hope it helps you.......

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