 Lena Bell

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:
$\mathbb{P}\left({E}_{1}{E}_{2}{E}_{3}\right)=1-\mathbb{P}\left({E}_{1}^{\prime }\cup {E}_{2}^{\prime }\cup {E}_{3}^{\prime }\right)=1-\left[3\mathbb{P}\left({E}_{1}^{\prime }\right)-3\mathbb{P}\left({E}_{1}^{\prime }{E}_{2}^{\prime }\right)+\mathbb{P}\left({E}_{1}^{\prime }{E}_{2}^{\prime }{E}_{3}^{\prime }\right)\right]$
(since $\mathbb{P}\left({E}_{i}^{\prime }\right)=\mathbb{P}\left({E}_{j}^{\prime }\right)$ and $\mathbb{P}\left({E}_{i}^{\prime }{E}_{j}^{\prime }\right)=\mathbb{P}\left({E}_{k}^{\prime }{E}_{m}^{\prime }\right)$).
Also: $\mathbb{P}\left({E}_{1}^{\prime }\right)=\frac{\left(\genfrac{}{}{0}{}{3}{0}\right)\left(\genfrac{}{}{0}{}{15}{6}\right)}{\left(\genfrac{}{}{0}{}{18}{6}\right)},$,
$\mathbb{P}\left({E}_{1}^{\prime }{E}_{2}^{\prime }\right)=\frac{\left(\genfrac{}{}{0}{}{3}{0}\right)\left(\genfrac{}{}{0}{}{15}{12}\right)}{\left(\genfrac{}{}{0}{}{18}{12}\right)},$
$\mathbb{P}\left({E}_{1}^{\prime }{E}_{2}^{\prime }{E}_{3}^{\prime }\right)=0.$
Do you see any flaw in the above argument? zlepljalz2

Expert

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 $\frac{12}{17}$ 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 $\frac{6}{16}=\frac{3}{8}$ 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 $\frac{36}{136}=\frac{9}{34}$. nidantasnu

Expert

Step 1
Your calculation is correct. We can confirm this by calculating the probability directly.
There are $\left(\genfrac{}{}{0}{}{18}{6}\right)\left(\genfrac{}{}{0}{}{12}{6}\right)\left(\genfrac{}{}{0}{}{6}{6}\right)$ 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 $\left(\genfrac{}{}{0}{}{15}{5}\right)\left(\genfrac{}{}{0}{}{10}{5}\right)\left(\genfrac{}{}{0}{}{5}{5}\right)$ ways to distribute the hard questions, so there are $3!\left(\genfrac{}{}{0}{}{15}{5}\right)\left(\genfrac{}{}{0}{}{10}{5}\right)\left(\genfrac{}{}{0}{}{5}{5}\right)$ 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 $\frac{3!\left(\genfrac{}{}{0}{}{15}{5}\right)\left(\genfrac{}{}{0}{}{10}{5}\right)\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{18}{6}\right)\left(\genfrac{}{}{0}{}{12}{6}\right)\left(\genfrac{}{}{0}{}{6}{6}\right)}$ which agrees with your answer.

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