Answer:
[tex]\sf \dfrac{12}{19}[/tex]
Step-by-step explanation:
To begin, add the column and row totals to the table (see attachment).
Let W = Wheat
Let C = With cheese
[tex]\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}[/tex]
[tex]\implies \sf P(W)=\dfrac{1900}{3125}=\dfrac{76}{125}[/tex]
[tex]\implies \sf P(C)=\dfrac{2000}{3125}=\dfrac{16}{25}[/tex]
[tex]\implies \sf P(W \cap C)=\dfrac{1200}{3125}=\dfrac{48}{125}[/tex]
As the events W and C are not independent, we can use the formula:
[tex]\sf P(C\: | \:W)=\dfrac{P(W \cap C)}{P(W)}[/tex]
Substituting the values into the formula:
[tex]\sf P(C\: | \:W)=\dfrac{\frac{48}{125}}{\frac{76}{125}}=\dfrac{48}{76}=\dfrac{12}{19}[/tex]