Events A and B are such that P(A)=1 2 , P(B)=7 12 and P(not A or not B)=1 4 . State whether A and B are independent?

It is given that P(A)=1 2 , P(B)=7 12 , and P(not A or not B)=1 4 (A' ')=1 4 ((A )' )=1 4 [A' '=(A )' ] 1-P(A )=1 4 (A )=3 4 ...(1) However, P(A) (B)=1 2 7 12 =7 24 ...(2) Here, 3 4 7 24 (A ) (A) (B) Therefore, A and B are independent events.

Events A and B are such that P(A)=1 2 , P(B)=7 12 and P(not A or …

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Events A and B are such that $\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=\frac{7}{12}\ \text{and}\ \text{P}(\text{not A or not B})=\frac{1}{4}.$ State whether A and B are independent?

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Answer

It is given that $\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=\frac{7}{12},\ \text{and}\ \text{P}(\text{not A or not B})=\frac{1}{4}$$\Rightarrow\text{P}(\text{A}'\cup\text{B}')=\frac{1}{4}$
$\Rightarrow\text{P}\Big((\text{A}\cap\text{B})'\Big)=\frac{1}{4}\ \ \ \ \ \Big[\text{A}'\cup\text{B}'=(\text{A}\cap\text{B})'\Big]$
$\Rightarrow1-\text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{3}{4}\ \ \ ...(1)$
$\text{However},\ \text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}\ \ \ ...(2)$
$\text{Here},\ \frac{3}{4}\neq\frac{7}{24}$
$\therefore\text{P}(\text{A}\cap\text{B})\neq\text{P}(\text{A})\cdot\text{P}(\text{B})$
Therefore, A and B are independent events.

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