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  • Let be two events such that <img alt="\mathrm{ P(\overline{A

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

Option: 1

mutually exclusive and independent
 


Option: 2

independent but not equally likely
 


Option: 3

equally likely but not independent
 


Option: 4

equally likely and mutually exclusive


Answers (1)

best_answer

Since, \mathrm{P(A \cap B)=\frac{1}{4} \neq 0}

Therefore, \mathrm{A \: and\: B} are not mutually exclusive events.

Now,\mathrm{P(\overline{A \cup B})=\frac{1}{6} }

\mathrm{\Rightarrow 1-P(A \cup B)=\frac{1}{6} }

\mathrm{\Rightarrow P(A \cup B)=\frac{5}{6} }

\mathrm{\Rightarrow\: P(A)+ P(B)-P(A \cap B)=\frac{5}{6} }

\mathrm{\Rightarrow P(B)=\frac{5}{6}+\frac{1}{4}-\frac{3}{4}=\frac{1}{3} }

Clearly, \mathrm{\quad P(A \cap B)=\frac{1}{4}=\frac{3}{4} \times \frac{1}{3}=P(A) \times P(B) }

So, \mathrm{A \: and\, B} are independent.

Also, \mathrm{P(A) \neq P(B) \Rightarrow A \: and \: B} are not equally likely.

Hence option 2 is correct.

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