nkmungila wrote:

In a class, every student is either American or European. Every student is wearing a blazer or a sweater. If a student is selected randomly, is the probability of the student being a European wearing a sweater greater than that of his/her being an American wearing a blazer?

(1) The probability that a randomly selected student is either an American or wearing a blazer is \(\frac{5}{6}\)

(2) The probability that a randomly selected student is either a European or wearing a sweater is \(\frac{3}{5}\)

Source:

http://www.expertsglobal.comWe need to find European wearing a sweater and American wearing a blazer

Statement 1: this implies that probability of European wearing a sweater \(= 1-\frac{5}{6}=\frac{1}{6}\)

but we know nothing about American wearing a blazer. Hence

InsufficientStatement 2: this implies that probability of American wearing a blazer \(= 1-\frac{3}{5}=\frac{2}{5}\)

but we know nothing about European wearing a sweater. Hence

InsufficientCombing 1 & 2 we get our two required values and \(\frac{2}{5}>\frac{1}{6}\). So we have a NO for our question stem. Hence

SufficientOption

C