Economist wrote:

A drawer contains 8 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

(1) The probability is less than 0.2 that the first sock is black.

(2) The probability is more than 0.8 that the first sock is white.

Number of Socks HAS to be an integer. Therefore, (1) & (2) are saying the same thing because

\(\frac{(Favorable Cases)}{(Total Number of Cases)} = Probability\)

=> Favorable Cases = Probability * Total Number of Cases

=> Number of Ways to pick a Black Sock in 1st Attempt = Prob. * 8

(1) The probability is less than 0.2 that the first sock is black. => Prob of 1at Black sock < 0.2

(2) The probability is more than 0.8 that the first sock is white. => Prob of 1st white > 0.8

=> Prob of 1st sock of any color other than white < 0.2

Therefore, Prob of 1st Black Sock < 0.2

Hence, the answer cannot be anything but D or E. D if the statement is sufficient otherwise E.

Now, Number of Ways to pick a Black Sock in 1st Attempt < 0.2 * 8 < 1.6

Hence, we have 0 or 1 Black sock.

Answer: D

_________________

I'd appreciate learning about the grammatical errors in my posts

Please hit Kudos If my Solution helps

My Daily Study Log to breach 750.

Thread of my notes to follow soon.