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given ( p(e) = 0.105 ), ( p(ecup f) = 0.351 ), ( p(ecap f) = 0.098 ), w…

Question

given ( p(e) = 0.105 ), ( p(ecup f) = 0.351 ), ( p(ecap f) = 0.098 ), what is ( p(f)=? ) note: enter your answer in the box below without rounding it.

Explanation:

Step1: Recall the formula for \(P(E\cup F)\)

The formula is \(P(E\cup F)=P(E)+P(F)-P(E\cap F)\).

Step2: Rearrange the formula to solve for \(P(F)\)

We get \(P(F)=P(E\cup F)+P(E\cap F)-P(E)\).

Step3: Substitute the given values

Substitute \(P(E) = 0.105\), \(P(E\cup F)=0.351\), and \(P(E\cap F)=0.098\) into the formula:
\(P(F)=0.351 + 0.098-0.105\).

Step4: Calculate the value

\(0.351+0.098 = 0.449\), then \(0.449-0.105=0.344\).

Answer:

\(0.344\)