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Given P(A)=0.48P(A)=0.48, P(B)=0.75P(B)=0.75 and P(A\cap B)=0.43P(A∩B)=0.43, find the value of P(A\cup B)P(A∪B), rounding to the nearest thousandth, if necessary.

Sagot :

We are given with the following :

  • [tex]{\sf{P(A)=0.48}}[/tex]
  • [tex]{\sf{P(B)=0.75}}[/tex]
  • [tex]{\sf{P(A\cap B)=0.43}}[/tex]

And we are asked to find [tex]{\bf{P(A\cup B)}}[/tex] . But before doing this , let's recall :

  • [tex]{\boxed{\bf{P(A\cup B)=P(A)+P(B)-P(A\cap B)}}}[/tex]

Now , putting the values

[tex]{:\implies \quad \sf P(A\cup B)=0.48+0.75-0.43}[/tex]

[tex]{:\implies \quad \sf P(A\cup B)=1.23-0.43}[/tex]

[tex]{:\implies \quad \sf P(A\cup B)=0.80=0.8}[/tex]

[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{P(A\cup B)=0.8}}}[/tex]

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