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Answer:
[tex]\text{1.}\\ \text{Let F denote the people who like football and C be the people who like }\\\text{cricket.}\\\text{Given, }\\\text{n(U) = 500}\\\text{n(F) = 250}\\\text{n(C) = 300}\\\text{n}(\overline{\text{F}\cup\text{C}})=25\\\text{n(F}\cap\text{C)}=x\text{ (Let)}\\[/tex]
[tex]\text{a. Solution: }\\\\\text{(See the posted image for venn-diagram.)}\\\\\text{n(F}\cup\text{C)}=\text{n(U) }-\text{n}(\overline{\text{F}\cup\text{C}})\\\text{or, }250-x+x+300-x=500-25\\\text{or, }550-x=475\\\text{or, }x=550-475\\\text{or, }x=75[/tex]
[tex]\text{b. }\\\text{Number of people who liked football only, n}_\circ(\text{F})=\text{n(F) }-\text{n}(\text{F}\cap\text{C})=250-75\\\therefore\ \text{n}_\circ(\text{F})=175\\\text{Number of people who liked cricket only, n}_\circ(\text{C})=\text{n(C) }-\text{n}(\text{F}\cap\text{C})=300-75\\\therefore\ \text{n}_\circ(\text{C})=225\\\text{So, number of people who liked only one game = }\text{n}_\circ(\text{C})+\text{n}_\circ(\text{F})=175+225\\\text{}\hspace{7.3cm}\text{= }400[/tex]
[tex]\text{c. }\\\text{Number of people who liked at most one game = n}_\circ(\text{F})+\text{n}_\circ(\text{C})+\text{n}(\overline{\text{F}\cup\text{C}})\\\text{}\hspace{7.3cm}\text{= }175+225+25=425[/tex]
[tex]\text{d.}\\\text{Number of people who liked at least one game = }\text{n(}\text{A}\cup\text{B})=250-x+x+300-x\\\text{}\hspace{9cm}\text{= }550-75=475[/tex]

