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:

$\mathbb{P}({E}_{1}{E}_{2}{E}_{3})=1-\mathbb{P}({E}_{1}^{\prime}\cup {E}_{2}^{\prime}\cup {E}_{3}^{\prime})=1-{\textstyle [}3\mathbb{P}({E}_{1}^{\prime})-3\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime})+\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime}{E}_{3}^{\prime}){\textstyle ]}$

(since $\mathbb{P}({E}_{i}^{\prime})=\mathbb{P}({E}_{j}^{\prime})$ and $\mathbb{P}({E}_{i}^{\prime}{E}_{j}^{\prime})=\mathbb{P}({E}_{k}^{\prime}{E}_{m}^{\prime})$).

Also: $\mathbb{P}({E}_{1}^{\prime})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{3}{0}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{15}{6}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{18}{6}{\textstyle )}},$,

$\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{3}{0}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{15}{12}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{18}{12}{\textstyle )}},$

$\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime}{E}_{3}^{\prime})=0.$

Do you see any flaw in the above argument?

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:

$\mathbb{P}({E}_{1}{E}_{2}{E}_{3})=1-\mathbb{P}({E}_{1}^{\prime}\cup {E}_{2}^{\prime}\cup {E}_{3}^{\prime})=1-{\textstyle [}3\mathbb{P}({E}_{1}^{\prime})-3\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime})+\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime}{E}_{3}^{\prime}){\textstyle ]}$

(since $\mathbb{P}({E}_{i}^{\prime})=\mathbb{P}({E}_{j}^{\prime})$ and $\mathbb{P}({E}_{i}^{\prime}{E}_{j}^{\prime})=\mathbb{P}({E}_{k}^{\prime}{E}_{m}^{\prime})$).

Also: $\mathbb{P}({E}_{1}^{\prime})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{3}{0}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{15}{6}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{18}{6}{\textstyle )}},$,

$\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{3}{0}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{15}{12}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{18}{12}{\textstyle )}},$

$\mathbb{P}({E}_{1}^{\prime}{E}_{2}^{\prime}{E}_{3}^{\prime})=0.$

Do you see any flaw in the above argument?

Answer & Explanation

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 $\frac{12}{17}$ 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 $\frac{36}{136}=\frac{9}{34}$.

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

2022-07-11Added 7 answers

Step 1

Your calculation is correct. We can confirm this by calculating the probability directly.

There are $(}\genfrac{}{}{0ex}{}{18}{6}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{12}{6}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{6}{6}{\textstyle )$ 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 $(}\genfrac{}{}{0ex}{}{15}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{10}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )$ ways to distribute the hard questions, so there are $3!{\textstyle (}\genfrac{}{}{0ex}{}{15}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{10}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )}$ 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 $\frac{3!{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{15}{5}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{10}{5}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )}}}{{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{18}{6}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{12}{6}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{6}{6}{\textstyle )}}}$ which agrees with your answer.

Your calculation is correct. We can confirm this by calculating the probability directly.

There are $(}\genfrac{}{}{0ex}{}{18}{6}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{12}{6}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{6}{6}{\textstyle )$ 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 $(}\genfrac{}{}{0ex}{}{15}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{10}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )$ ways to distribute the hard questions, so there are $3!{\textstyle (}\genfrac{}{}{0ex}{}{15}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{10}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )}$ 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 $\frac{3!{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{15}{5}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{10}{5}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )}}}{{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{18}{6}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{12}{6}{\textstyle )}}{\displaystyle {\textstyle (}\genfrac{}{}{0ex}{}{6}{6}{\textstyle )}}}$ which agrees with your answer.

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