At a medical research lab, nine doctors are conducting multiple clinic

Two approaches.

First: direct probability

There are 6 members in the first group and 3 in the second.

The probability of a doctor being selected from the first group is

6/9=2/3

Since there are 8 remaining doctors to choose from, but the first doctor's partner is disqualified, the probability of picking a non partner is

7/8

So the probability of two doctors being picked that aren't partners where 1 doctor is picked from the first group is

2/3*7/8 = 14/24

There is a complementary chance of 1/3 of picking the doctor from the second group of 3

With 8 doctors again remaining, but now 2 doctors are disqualified, the probability of picking a non partner is

6/8


So the probability of 2 doctors being picked that aren't partners where 1 doctor is picked from the second group is


1/3*6/8 = 6/24

The overall probability is then

14/24 + 6/24 = 20/24 = 10/12


Using combinations

The number of ways to pick a doctor from the first group

6

The number of ways to then pick a non partner

7

Total ways 6*7/2 = 21. Divide by 2 because order doesn't matter.

Number of ways to pick a doctor from the second group

3

Number of ways to pick a non partner

6

Total ways 3*6/2 = 9, again dividing by 2 because order doesn't matter

So total ways is

21+9 = 30

To compute probability, need total ways 2 doctors can be chosen from 9

9!/2!7! = 36

Probability: 30/36 = 10/12

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In this part, won't it be more than 3 ways, for example - PQ, PR, PS, PT, PU, QR, QS, QT, QU,...?

Am I missing something here?

Bunuel, when will the OA be published?

Let's say that the 9 doctors are ABCDEFGHI.

Six of the doctors are working on a clinical trial with exactly one other doctor.
Let's say that A and B are working together, that C and D are working together, and that E and F are working together.

Three doctors are working on a clinical trial with exactly two other doctors.
Let's say that G, H and I are all working with one another.

Total number of doctors pairs that are working together = 6 (AB, CD, EF, GH, GI, and HI)
Total number of pairs that can be formed from 9 doctors = 9C2 = \(= \frac{9*8}{2*1} = 36\)
P(random pair is working together) = 6/36 = 1/6
P(random pair is NOT working together = 1 - 1/6 = 5/6 = 10/12

You're picking two that work together, not any two, which would be a combination.

There are only 3 sets of 2 that work together.

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