Answer:
By definition, \[P(A/B)=\frac{P(A\cap B)}{P(B)}\] ?..(i) Also, \[A\cap B\subset B\] \[[\because \,\,\,A\cap B)\,\,is\,\,a\,\,subset\,\,of\,\,B]\] \[\Rightarrow \] \[P(A\cap B)\le P(B)\] \[\Rightarrow \] \[\frac{P(A\cap B)}{P(B)}\le 1\] ?.(ii) Again, \[P(A\cap B)\ge 0,\,and\,B\ne \phi \] \[\therefore \] \[P(B)>0\] \[\therefore \] \[\frac{P(A\cap B)}{P(B)}\ge 0\] ?(iii) From Eqs. (ii) and (iii) we have \[0\le \frac{P(A\cap B)}{P(B)}\le 1\] Hence, \[0\le P(A/B)\le 1\] [from Eq. (i)] Hence proved.
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