Answer to Question #204424 in Statistics and Probability for jay

Question

. George estimates that there is a 30% chance of rain the next day if he waters the lawn, a 40% chance if he washes the car, and a 50% chance if he plans a trip to the beach. Assuming George’s estimates are accurate, what are the odds a) in favour of rain tomorrow if he waters the lawn? b) in favour of rain tomorrow if he washes the car? c) against rain tomorrow if he plans a trip to the beach?

Answer 1

$Given \\ chance \space of \space rain \space the \space next \space Day \space if \space he \space waters \space the \space lawn=Pr(A)=30\%\\ chance \space of \space rain \space the \space next \space Day \space if \space he \space washes \space the \space car=Pr(B)=40\%\\ chance \space of \space rain \space the \space next \space Day \space if \space he \space plan \space the \space trip=Pr(C)=50\%\\ Now \space we \space know \space that, \\Pr(X)+Pr( \bar{X })=1 \\ Now \\ a)odds \space in \space favour \space of \space rain \space tomorrow \space if \space he \space waters \space the \space lawn? Pr(A)=30\%\\ Pr( \bar{A })=70\%\\ Then \\ odds \space in \space favour \space of \space rain \space tomorrow \space if \space he \space waters \space the \space lawn=30:70 \\ =3:7 \\ b)odds \space in \space favour \space of \space rain \space tomorrow \space if \space he \space washes \space the \space car? Pr(B)=40\%\\ Pr( \bar{B})=60\%\\ Then \\ odds \space in \space favour \space of \space rain \space tomorrow \space if \space he \space Washes \space the \space car=40:60 \\ =4:6 \\ c)odds \space in \space against \space of \space rain \space tomorrow \space if \space he \space plans \space the \space trip? Pr(C)=50\%\\ Pr( \bar{C })=50\%\\ Then \\ odds \space in \space against \space of \space rain \space tomorrow \space if \space he \space plan \space the \space trip= Pr( \bar{C }):Pr(C)\\=50:50 \\ =1:1 \\$