एक न्यध्या पासे को उछाला गया हैं। घटनाओं E={1,3,5} , F={2,3} ,G={2,3,4,5} (ii) `{P(E//G)` और `P(G//E)`
(iii) `P(E cap F//G)` और `P(E cap F//G)`

Explanation

यहाँ, प्रतिदर्श समषिट `s=(1,2,3,4,5,6)`
दिया हैं, E={1,3,5}, F={2,3}
तथा G={2,3,4,5}
`rArr E cap F = {3}, E cap G= {3,5}`
`E cap F = {1,2,3,5}, (E cup F) cap G={2,3,5}`
तथा `(E cap F) = {3}, E cap G = {3,5}`
`E cup F = {1,2,3,5}, {E cup F} cap G ={2,3,5}`
`rArr n(S) = 6, n(E ) =3` तथा `n(E cap G) =2, n(E cup F) cap G=1`
अतः `P(E ) =3/6=1/2, P(F ) =2/6 =1/3, P[G] =4/6=2/3`
`P(E cap F)=1/6, P(E cap G) =2/6=1/3`
`P[(E cap F) cap G]=3/6=1/2 ` तथा `[P(E cap F) cap G]=1/6`
(i) `P(E/F) = (P(E cap F))/(P(F)) = (1//6)/(1//3) =1/2`
तथा `P(F/E) = (P(F cap E))/(P(E)) = (1//6)/(1//2) = 1/3`
(ii) `P(E/G) = (P(E cap G))/(P(G)) = (1//3)/(2//3) =1/2`
`P(G/E) = (P(G cap E))/(P(E)) = (1//3)/(1//2) =2/3`
(iii) `P(E cap F)/(G) = P[(E cap F) cup G]/(P(G)) = (1//6)/(2//3)=1/4`