If P A -65- P B -80- then P A - B lies in the intervalA- -4- -70-B- -45- -65-C- -30- -80-D- -35- -75-

P(A)=0.65, P(B)=0.80 Condition 1: P(A∩ B)≤min(P(A),P(B)) P(A∩ B)≤0.65...(1) Condition 2: P(A∩ B)=P(A)+P(B) P(A∪ B) As we know, P(A∪ B)≤ 1 ⇒ P(A∩ B)≥ P(A)+P(B) 1 ...

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Standard XII

Mathematics

Axiomatic Approach

If P A =65, P...

Question

If $P (A) = 65, P (B) = 80$ , then $P (A \cap B)$ lies in the interval

[.30, .80]

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[.35, .75]

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[.4, .70]

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[.45, .65]

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Solution

The correct option is D $[.45, .65]$
$P (A) = 0.65$ , $P (B) = 0.80$
Condition 1: $P (A \cap B) \leq min (P (A), P (B))$
$P (A \cap B) \leq 0.65... (1)$
Condition 2:
$P (A \cap B) = P (A) + P (B) - P (A \cup B)$
As we know, $P (A \cup B) \leq 1$
$\Rightarrow P (A \cap B) \geq P (A) + P (B) - 1$
$\geq 0.65 + 0.80 - 1$
$\Rightarrow P (A \cap B) \geq 0.45... (2)$
From $(1)$ and $(2)$
$0.45 \leq P (A \cap B) \leq 0.65$

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If P(A)=65,P(B)=80, then P(A∩B) lies in the interval

The correct option is D [.45,.65]P(A)=0.65, P(B)=0.80 Condition 1: P(A∩B)≤min(P(A),P(B)) P(A∩B)≤0.65...(1) Condition 2: P(A∩B)=P(A)+P(B)−P(A∪B) As we know, P(A∪B)≤1 ⇒P(A∩B)≥P(A)+P(B)−1 ≥0.65+0.80−1 ⇒P(A∩B)≥0.45...(2) From (1) and (2) 0.45≤P(A∩B)≤0.65

If $P (A) = 65, P (B) = 80$ , then $P (A \cap B)$ lies in the interval