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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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If $ P(A)=0.65\: P(B)=0.80$ then $ P(A \cap B)$ lies in the interval

$\begin {array} {1 1} (A)\;[.30,80] & \quad (B)\;[0.35, 0.75] \\ (C)\;[0.4, 70] & \quad (D)\;[ 0.45, 0.65] \end {array}$

 

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1 Answer

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$ P(A \cap B) \leq min. {P(A),P(B)}\: =\: min.{0.65, 0.80}$
$= 0.65$
$ \therefore P(A \cap B ) \leq 0.65$
Also $ P(A \cap B)= P(A)+P(B)-P( A \cup B) \geq 0.65 + 0.80-1$
$( \because P(A \cup B) \leq 1 )$
$ = 0.45 \: \: \: \: \therefore -P ( A \cup B ) \geq -1$
$ \therefore 0.45 \leq P( A \cap B ) \leq 0.65$
Ans : (D)
answered Jan 6, 2014 by thanvigandhi_1
 

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