Question

A coffee roaster has 12 rare coffee beans with intensity scores ranked from 1 (mildest) to 12 (strongest). You choose 7 beans at random and line them up from mildest to strongest: \(C_1<C_2<C_3<C_4<C_5<C_6<C_7\). What is the probability that the third bean \(C_3\) has an intensity score of exactly 4?

• A. \(\dfrac{5}{18}\)
• B. \(\dfrac{35}{132}\)
• C. \(\dfrac{21}{44}\)
• D. \(\dfrac{1}{4}\)

Hint:

Whenever a particular ordered position is fixed, divide the remaining elements into:
• numbers smaller than the fixed value,
• numbers larger than the fixed value. Then apply combinations separately.

Answer 1

The Correct Option is B