Question

At a medical research lab, nine doctors are conducting multiple clinical trials. Six of the doctors are working on a clinical trial with exactly one other doctor and three doctors are working on a clinical trial with exactly two other doctors. If two doctors are selected at random from the lab, what is the probability that those two doctors are NOT working together on a clinical trial?

• A. 1/12
• B. 2/12
• C. 5/12
• D. 7/12
• E. 10/12

Hint:

For probability questions asking for the chance of something "not" happening, it's often simpler to calculate the probability of it happening and subtract that from 1. This method, known as using the complement, can prevent complex counting.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This is a probability problem that requires using combinations. The strategy is to find the total number of possible pairs of doctors, then find the number of pairs that are working together, and finally calculate the probability of picking a pair that is NOT working together.
Step 2: Key Formula or Approach:
The number of ways to choose a group of \(k\) items from a set of \(n\) items is given by the combination formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
The probability of an event is \( P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \).
The probability of an event NOT happening is \( P(\text{not A}) = 1 - P(\text{A}) \).
Step 3: Detailed Explanation:
1. Determine the structure of the working groups:
- "Six of the doctors are working on a clinical trial with exactly one other doctor." This means these 6 doctors form 3 pairs. Let's call them (D1, D2), (D3, D4), (D5, D6). The number of working pairs here is 3.
- "three doctors are working on a clinical trial with exactly two other doctors." This means these 3 doctors are in a single group of three. Let's call them (D7, D8, D9). The pairs working together in this group are (D7, D8), (D7, D9), and (D8, D9). The number of working pairs here is \( \binom{3}{2} = \frac{3 \times 2}{2} = 3 \).
2. Calculate the total number of pairs working together:
Total "working" pairs = (Pairs from the groups of two) + (Pairs from the group of three) = \( 3 + 3 = 6 \).
3. Calculate the total number of possible pairs of doctors:
We are selecting 2 doctors from a total of 9. The total number of possible pairs is:
\[ \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36 \] 4. Calculate the probability of selecting two doctors who ARE working together:
\[ P(\text{together}) = \frac{\text{Number of working pairs}}{\text{Total possible pairs}} = \frac{6}{36} = \frac{1}{6} \] 5. Calculate the probability of selecting two doctors who are NOT working together:
The question asks for the probability that the two selected doctors are NOT working together. We can find this using the complement rule:
\[ P(\text{NOT together}) = 1 - P(\text{together}) \] \[ P(\text{NOT together}) = 1 - \frac{1}{6} = \frac{5}{6} \] 6. Match the result with the options:
The options are given in terms of a denominator of 12. Let's convert our answer:
\[ \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12} \] Step 4: Final Answer:
The probability that the two selected doctors are NOT working together is 10/12.