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  • Let A and B two events such that  <img alt="P(\overline{A\cup B})=\frac{1}{6},\:\:P(A\cap B )=\frac{1}{4}\:\:and\:\:P (\overline{A})=\frac{1}{4}" src="https://learn.careers360.com/latex-image/

Let A and B two events such that  P(\overline{A\cup B})=\frac{1}{6},\:\:P(A\cap B )=\frac{1}{4}\:\:and\:\:P (\overline{A})=\frac{1}{4}   where \bar{A} stands for the complement of the event A. Then the events A and B are :

Option: 1

  independent but not equally likely.
 


Option: 2

independent and equally likely.


Option: 3

mutually exclusive and independent.

 


Option: 4

equally likely but not independent.


Answers (1)

best_answer

As learnt

Addition Theorem of Probability -

P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )

in general:

P\left ( A_{1}\cup A_{2}\cup A_{3}\cdots A_{n} \right )=\sum_{i=1 }^{n}P\left ( A_{i} \right )-\sum_{i< j}^{n}P\left ( A_{i}\cap A_{j} \right )+\sum_{i< j< k}^{n} P\left ( A_{i}\cap A_{j}\cap A_{k} \right )-\cdots -\left ( -1 \right )^{n-1}P\left ( A_{1}\cap A_{2}\cap A_{3}\cdots \cap A_{n} \right )

-

 

 and

 

Independent events -

If A and B are independent events then A and \overline{B} as well as \overline{A} and B are independent events. 

- wherein

 

   P\left ( A\cup B \right )=\frac{5}{6} \: , \: P(A\cap B)=\frac{1}{4}

P(A)= \frac{3}{4} \: \: Let (B)=x

P(A\cup B)= \frac{5}{6}=\frac{3}{4}+x-\frac{1}{4}

x=\frac{5}{6}-\frac{1}{2}=\frac{1}{3}

P(A\cap B)+P(A)\cdot P(B)=\frac{1}{4} \: \: independent \: events

Also, P(A)\neq P(B) \ not \: equally \: likely

Posted by

avinash.dongre

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