Each group gets 6 out of 18 questions (3 easy, 15 hard), probability all groups...

Lena Bell

Lena Bell

Answered

2022-07-09

Each group gets 6 out of 18 questions (3 easy, 15 hard), probability all groups have one easy question
In a course, students form three groups. Each group is assigned 6 questions, chosen at random, without replacement, from a set of 18 questions. From these 18 questions, 3 are easy and 15 are hard. What is the probability all groups have one easy question?
Attempt. Let E i be the event group i has at least one easy question. The desired event is E 1 E 2 E 3 and using complements and inclusion/exclusion:
P ( E 1 E 2 E 3 ) = 1 P ( E 1 E 2 E 3 ) = 1 [ 3 P ( E 1 ) 3 P ( E 1 E 2 ) + P ( E 1 E 2 E 3 ) ]
(since P ( E i ) = P ( E j ) and P ( E i E j ) = P ( E k E m )).
Also: P ( E 1 ) = ( 3 0 ) ( 15 6 ) ( 18 6 ) ,,
P ( E 1 E 2 ) = ( 3 0 ) ( 15 12 ) ( 18 12 ) ,
P ( E 1 E 2 E 3 ) = 0.
Do you see any flaw in the above argument?

Answer & Explanation

zlepljalz2

zlepljalz2

Expert

2022-07-10Added 22 answers

Step 1
There's another way to approach this. Instead of assigning questions to groups, assign groups to questions. In other words, look at which group each question is assigned to.
Step 2
Assign the first easy question to a group arbitrarily. For this to be a "good" assignment, there is a probability of 12 17 that the second easy question will be assigned to a different group. That's because there remain 17 vacant spaces of which 12 are good. If that's the case, there is a 6 16 = 3 8 probability that the third easy question will be assigned to the third group because now there are 16 remaining vacant spaces of which 6 are good. Thus, the probability of a "good" assignment is 36 136 = 9 34 .
nidantasnu

nidantasnu

Expert

2022-07-11Added 7 answers

Step 1
Your calculation is correct. We can confirm this by calculating the probability directly.
There are ( 18 6 ) ( 12 6 ) ( 6 6 ) ways to distribute the cards to three labeled groups.
If each group receives one easy question, then there are 3! ways to distribute the easy questions and ( 15 5 ) ( 10 5 ) ( 5 5 ) ways to distribute the hard questions, so there are 3 ! ( 15 5 ) ( 10 5 ) ( 5 5 ) ways to distribute the questions to three labeled groups so that each group receives one easy question.
Step 2
Hence, the probability that each group receives one easy question is 3 ! ( 15 5 ) ( 10 5 ) ( 5 5 ) ( 18 6 ) ( 12 6 ) ( 6 6 ) which agrees with your answer.

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