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Given that  \mathrm{x \in[0,1]}  and\mathrm{ y \in[0,1]} . Let \mathrm{A}  be the event \mathrm{(x, y)} satisfying \mathrm{y^2 \leq x}  and \mathrm{B}  be the event of \mathrm{ (x, y)} satisfying \mathrm{x^2 \leq y}. Then 

Option: 1

\mathrm{P(A \cap B)=\frac{1}{3}}


Option: 2

A, B are exhaustive


Option: 3

A, B are mutually exclusive
 


Option: 4

A, B are independent


Answers (1)

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\mathrm{A}=  the event of \mathrm{(x, y)} belonging to the area OTQPO \mathrm{B}= the event of  \mathrm{(x, y)} belonging to the area OSQRO

\mathrm{\begin{aligned} P(A) & =\frac{\operatorname{ar}(\text { OTQPO })}{\operatorname{ar}(\text { OPQRO) }}=\frac{\int_0^1 \sqrt{x d x}}{1 \times 1} \\ & =\left[\frac{2}{3} x^{3 / 2}\right]_0^1=\frac{2}{3} . \end{aligned}}

\mathrm{P(B)=\frac{\operatorname{ar}(\mathrm{OSQRO})}{\operatorname{ar}(\mathrm{OPQRO})}=\frac{\int_0^1 \sqrt{y d y}}{1 \times 1}=\frac{2}{3}}

\mathrm{P(A \cap B)=\frac{\operatorname{ar}(\mathrm{OTQS})}{\operatorname{ar}(\mathrm{OPQRO})}=\frac{\int_0^1 \sqrt{x d x-\int_0^1 x^2 d x}}{1 \times 1}=\frac{2}{3}-\frac{1}{3}=\frac{1}{3} .}

\mathrm{P(A)+P(B)=\frac{2}{3}+\frac{2}{3} \neq 1} So, \mathrm{A}  and \mathrm{B}  are not exhaustive.
\mathrm{P(A) \cdot P(B)=\frac{2}{3} \cdot \frac{2}{3}=\frac{4}{9} \neq P(A \cap B)}. So, A and B are not independent.
\mathrm{P(A \cup B)=1, P(A)+P(B)=\frac{2}{3}+\frac{2}{3} \neq P(A \cup B)} So \mathrm{A} and \mathrm{B} are not mutually exclusive.

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Divya Prakash Singh

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