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

We need to determine the probability that the third bean \(C_3\) has an intensity score of exactly 4 when 7 beans are chosen at random from 12 beans with scores ranging from 1 to 12.

Step 1: Identify Positions for Bean \(C_3\). 
The third bean \(C_3\) must have a score of 4, which means it already takes a fixed position among the selected 7 beans.

Step 2: Choose Remaining Beans. 
With \(C_3\) being 4, to ensure correct ordering from mildest to strongest, we choose:

• Two beans from intensities 1 to 3.
• Four beans from intensities 5 to 12.

The presence of these limitations leads to the following binomial selections:

• Number of ways to choose 2 beans from the set \(\{1, 2, 3\}\): \(\binom{3}{2} = 3\).
• Number of ways to choose 4 beans from the set \(\{5, 6, \ldots, 12\}\): \(\binom{8}{4} = 70\).

Step 3: Total Number of Favorable Outcomes. 
Multiply the individual combinations:

\(3 \times 70 = 210\)

Step 4: Calculate Total Possible Outcomes. 
Number of ways to choose any 7 beans out of 12:

\(\binom{12}{7} = 792\)

Step 5: Calculate Probability.
Using the probability formula:

\(\frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{210}{792}\)

Simplify the fraction:

\(\frac{210}{792} = \frac{35}{132}\)

Therefore, the probability that the third bean \(C_3\) has an intensity score of exactly 4 is:

Correct Answer: \(\frac{35}{132}\)