COMEDK · Maths · 36. Probability
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 \(\dfrac{1}{4}\)
- C \(\dfrac{21}{44}\)
- D \(\dfrac{35}{132}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{35}{132}\)
Step-by-step Solution
Detailed explanation
The total number of ways to select 7 beans from 12 is \(^{12}C_{7}\). For the third bean \(C_{3}\) to have an intensity score of exactly 4, the selection must satisfy the following conditions: Two beans must be chosen from the 3 beans with scores less than 4 (scores 1, 2, 3).…
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