Bunuel 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 \(\frac{3}{5}\)
There are 2 categories of people and within each of them, there are 2 sub categories.
In these questions, use 2 by 2 matrix. Let the four categories be: \(E_b, E_s,A_s,A_b\) equal to w, y, z and x respectively as shown below.
2 by 2 matrix.
\(…………………E…………A………….Total\)
\(B\)………….…w…………x……….w+x
\(S\)……….……..y…………z……..y+z
Total……....w+y………..x+z.….w+x+y+z
We are looking for whether \(\frac{E_s}{t}>\frac{A_b}{t}\) or \(\frac{y}{t}>\frac{x}{t}\)?
(1) The probability that a randomly selected student is either an American or wearing a blazer is \(\frac{5}{6}\)
This tells us that \(\frac{w+x+z}{t}=\frac{5}{6}\).
\(1-\frac{w+x+z}{t}=1-\frac{5}{6}\)
So, \(P(E_s)=\frac{y}{t}=\frac{1}{6}\)
Nothing about x/t
(2) The probability that a randomly selected student is either a European or wearing a sweater is \(\frac{3}{5}\)
This tells us that \(\frac{w+y+z}{t}=\frac{3}{5}\).
\(1-\frac{w+y+z}{t}=1-\frac{3}{5}\)
So, \(P(A_b)=\frac{x}{t}=\frac{2}{5}\)
Nothing about y/t
Combined
We can clearly say 2/5>1/6 or x/t>y/t
Sufficient
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