Question 1018171
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       A(0.25)       B(0.15)     P(Both)    =  0.0375  [P(A) * P(B)]
                 Not B(0.85)     P(A only)  =  0.2125  [P(A) * P(Not B)]

   Not A(0.75)       B(0.15)     P(B only)  =  0.1125  [P(Not A) * P(B)]
                 Not B(0.85)     P(Neither) =  0.6375  [P(Not A) * P(Not B)]


Part a, b, and c are read directly from the table.

Part d, "Not Both" is 1 minus P(Both) which is the same as P(A only) plus P(B only) plus P(Neither).

Part e, "Not at least one", is the same as P(Neither); see table.
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John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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