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

Concept: When objects are selected and arranged automatically in increasing order, combinations are used instead of permutations. The probability is calculated as: \[ \text{Probability}= \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} \]

Step 1: Finding the total number of possible selections. We must choose 7 beans from 12 beans. Therefore, the total number of selections is: \[ ^{12}C_7 \] Using the combination formula: \[ ^{12}C_7= \frac{12!}{7!5!} \] \[ ^{12}C_7= \frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1} \] \[ ^{12}C_7=792 \] Hence, total outcomes are: \[ 792 \]

Step 2: Understanding the condition \(C_3=4\). The selected beans are arranged in increasing order: \[ C_1<C_2<C_3<C_4<C_5<C_6<C_7 \] If the third bean equals 4, then:
• Exactly two selected beans must be smaller than 4.
• Those two beans must come from \(\{1,2,3\}\).
• Exactly four selected beans must be greater than 4.
• Those four beans must come from \(\{5,6,7,8,9,10,11,12\}\).

Step 3: Finding favorable outcomes. Number of ways to choose 2 beans from \(\{1,2,3\}\): \[ ^{3}C_2=3 \] Number of ways to choose 4 beans from the remaining 8 larger beans: \[ ^{8}C_4=70 \] Therefore, favorable outcomes are: \[ 3\times70=210 \]

Step 4: Calculating the required probability. \[ P(C_3=4)=\frac{210}{792} \] Dividing numerator and denominator by 6: \[ P(C_3=4)=\frac{35}{132} \]