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3 years ago

14

HEllpplppll I’ll give u a star

2 answers:

Firdavs [7]3 years ago

8 0

That’s not a function, hope this helps:)

denpristay [2]3 years ago

7 0

Answer:

its not a function :)

Step-by-step explanation:

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(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0

Goshia [24]

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

$P(A)+P(B)+P(A\cap B)=P(A\cup B)$

We know that

Maximum value of $P(A\cup B)$ =1 and minimum value of $P(A\cup B)$ =0

$0\leq P(A\cup B )\leq 1$

$0\leq P(A)+P(B)-P(A\cap B)\leq 1$

$0\leq 0.4+0.7-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)$

$P(A\cap B)\leq 1.1$

It is not possible that $P(A\cap B)$ is equal to 1.1

$1.1-P(A\cap B)\leq 1$

$-P(A\cap B)\leq 1-1.1=-0.1$

Multiply by (-1) on both sides

$P(A\cap B)\geq 0.1$

Again, $P(A\cup B)\geq P(B)$

$0.4+0.7-P(A\cap B)\geq 0.7$

$1.1-P(A\cap B)\geq 0.7$

$-P(A\cap B)\geq -1.1+0.7=-0.4$

Multiply by (-1) on both sides

$P(A\cap B)\leq 0.4$

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3 years ago

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Step-by-step explanation:

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