Answer:
The required prove is shown below:
Step-by-step explanation:
Consider the provided information
P(A) = 0.4 and P(B) = 0.7
We know that:
[tex]P(A\cup B)=P(A)+P(B)-P(AB)[/tex]
The maximum value of [tex]P(A\cup B)[/tex] is less than or equal to 1.
[tex]P(A\cup B)\leq 1[/tex]
[tex]P(A)+P(B)-P(AB)\leq 1[/tex]
[tex]0.4+0.7-P(AB)\leq 1[/tex]
[tex]1.1-P(AB)\leq 1[/tex]
[tex]P(AB)\geq 0.1[/tex]......(1)
Also [tex]P(A\cup B)\geq P(B)[/tex]
Now substitute the respective values.
[tex]P(A)+P(B)-P(AB)\geq P(B)[/tex]
[tex]0.4-P(AB)\geq 0[/tex]
[tex]P(AB)\leq 0.4[/tex].....(2)
Now from equation 1 and 2 we get.
[tex]0.1\leq P(AB)\leq 0.4[/tex]
Hence, proved
