Each group gets 6 out of 18 questions (3 easy, 15 hard), probability all groups have one easy questionIn 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?

Question

Each group gets 6 out of 18 questions (3 easy, 15 hard), probability all groups have one easy questionIn 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?

zlepljalz2 · Accepted Answer

Step 1There&#039;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 2Assign the first easy question to a group arbitrarily. For this to be a &quot;good&quot; assignment, there is a probability of       12    17   that the second easy question will be assigned to a different group. That&#039;s because there remain 17 vacant spaces of which 12 are good. If that&#039;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 &quot;good&quot; assignment is       36    136    =      9    34  .

nidantasnu · Accepted Answer

Step 1Your 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 2Hence, 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.