if a and b are events such that p(AuB)=5/6,p(A^)=(1/4),p(B)=1/3 then A and B are ​

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if a and b are events such that p(AuB)=5/6,p(A^)=(1/4),p(B)=1/3 then A and B are

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Serenity 3 weeks 2021-08-20T16:43:05+00:00 2 Answers 0 views 0

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    0
    2021-08-20T16:44:15+00:00

    Given,

    \longrightarrow\sf{P(A')=\dfrac{1}{4}\implies P(A)=1-\dfrac{1}{4}=\bold{\dfrac{3}{4}}}

    \longrightarrow\sf{P(B)=\dfrac{1}{3}}

    \longrightarrow\sf{P(A\cup B)=\bold{\dfrac{5}{6}}}

    We know ,

    \boxed{\bold{\sf{P(A\cap B)=P(A)+P(B)-P(A\cup B)}}}

    \implies\sf{P(A \cap B)=\dfrac{3}{4}+\dfrac{1}{3}-\dfrac{5}{6}}

    \implies\sf{P(A \cap B)=\dfrac{9}{12}+\dfrac{4}{12}-\dfrac{10}{12}}

    \implies\sf{P(A \cap B)=\dfrac{9+4-10}{12}}

    \implies\sf{P(A \cap B)=\dfrac{3}{12}}

    \implies\sf{P(A \cap B)=\dfrac{1}{4}}

    ______________________________

    \sf{P(A)\times P(B)=\dfrac{3}{4}\times \dfrac{1}{3}}

    \implies\sf{P(A)\times P(B)=\dfrac{3}{12}}

    \implies\sf{P(A)\times P(B)=\dfrac{1}{4}}

    \implies\sf{P(A)\times P(B)=P(A\cap B)}

    ∴ A and B are independent but not equally likely.

    0
    2021-08-20T16:44:39+00:00

    Answer:

    They are independent

    P(A) × P(B)= P(A intersection B)

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