A used car dealer has 30 cars and 10 of them are lemons (i.e.~ mechanically faulty used cars), and you don't know which is which. If you buy 3 cars, what is the probability that you will get at least one lemon?

Answer: Hence, the probability that he will get at least one lemon is 0.70.Step-by-step explanation:Since we have given thatNumber of cars = 30Number of lemon cars = 10Number of other than lemon cars = 30-10 = 20According to question, he bought 3 cars, we need to find the probability that you will get at least one lemon.So, P(X≤1)=1-P(X=0)=1-P(no lemon)Here, P(no lemon ) is given by[tex]\dfrac{20}{30}\times \dfrac{20}{30}\times \dfrac{20}{30}=(\dfrac{20}{30})^3[/tex]so, it becomes,[tex]P(X\geq 1)=1-(\dfrac{20}{30})^3=1-(0.67)^3=0.70[/tex]Hence, the probability that he will get at least one lemon is 0.70.