Question

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)​

Answer 1

GIVEN :–

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• P(AUB) = ¾

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

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

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TO FIND :–

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

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SOLUTION :–

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• We know that –

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

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• So that –

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

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

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

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$\rule{220}{2}$

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

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

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

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

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

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$\rule{220}{2}$

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▪︎ Now Let's find P(B) –

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

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• Put the values –

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

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

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

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

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

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

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$\rule{220}{2}$

Answer 2

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.......