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This is a solved example in Introduction to Probability by Tsitsiklis, page 26, example 1.11 A Class consists of 4 graduate and 12 undergraduate students is randomly divided into 4 groups of 4. What is the probability that each includes a graduate student?

I understand the combinatorial way to do this problem using multinomial theorem. I am here interested in the conditional probability approach.

Solution: Let us denote the four graduate students by 1,2,3,4 and consider the events $A_1$ ={students 1 and 2 are in different groups} $A_2$ = {students 1,2 and 3 are in different groups} $A_3$ ={students 1,2,3 and 4 are in different groups} We will calculate $P(A_3) = P(A_1 \cap A_2\cap A_3)= P(A_1)P(A_2|A_1)P(A_3|A_1\cap A_2)$ We have $P(A_1) =\frac{12}{15}$ since there are 12 students slot in groups other than the one of the student 1, and there are 15 student slots overall, excluding student 1. Similarly $P(A_2|A_1) =\frac{8}{14}$ and $P(A_3|A_1\cap A_2) =\frac{4}{13}$ and so $P(A_3) = \frac{12}{15} \times \frac{8}{14} \times \frac{4}{13}$

The question I have is how did he get $P(A_1)=\frac{12}{15}$ ? I understand that there are 4 graduate students who need to be in 4 different groups and along with 3 undergraduate students. I understand the reason for finding $P(A_1 \cap A_2 \cap A_3)$. What I don't understand is how he got $P(A_1) = \frac{12}{15}$. Do we have two groups of 4 students each with two graduate students in two different groups ? Can someone explain to me the size of the two groups we have so formed with $A_1$ and similarly with $A_2$ and $A_3$? If i have calculate $P(A_1),P(A_2|A_1),P(A_3|A_1\cap A_2)$ probabilities using counting how will i come to these numbers ? Thank you.

user115368
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    To compute $P(A_1)$: Student $1$ is in group $X$. There are then $15$ spots available for student $2$, exactly $12$ of which are outside $X$. The claim follows. – lulu Dec 16 '16 at 00:26
  • Do the 4 groups have to contain 4 students each ? – WW1 Dec 16 '16 at 00:33
  • Yes, Each group has to have one graduate student and three undergraduate students. That's why I am confused when the solution talks about $P(A_1)$ where the two graduate students are in different groups. I am not sure how in the calculation $\frac{12}{15}$ he manages the size of the two groups ? – user115368 Dec 16 '16 at 00:52
  • I was here for the same explanation, and noticed that @lulu had done it in the first comment. But what is implicit in this probability, is the fact that it is also conditional. – Mariusz Popieluch Oct 26 '23 at 14:44

2 Answers2

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Imagine that you are randomly placing students into 4 different groups. You "draw" (or randomly place) graduate student 1 first, and you put him or her in group x. You then have to place the remaining 15 students into the 4 groups (where group x has 3 remaining spots, and the other groups have 4 remaining spots).

What is the probability that any single student (i.e. graduate student 2) will also be in group x? Since there are 3 spots left in group x and a total of 15 students to choose from, $ P = \frac{3}{15} $. However, you are interested in the probability that any single student (i.e. graduate student 2) will not be in group x, which amounts to $ P(A_1) = 1 - P = 1 - \frac{3}{15} = \frac{15-3}{15} = \frac{12}{15} $

The order of the draws is not important (i.e. whether you choose student 4 or student 8 first doesn't change anything), but it often helps to think about these processes in such a way.

You can use a similar approach/method to get $ P(A_2|A_1) $ and $ P(A_3|A_1\cap A_2) $...

Michael R
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  • Hello thank you for your clear answer, now I understood the math behind P(A1), but I couldn't apply the same logic on P(A2). If we randomly pick 2 grad students and put them in group x than the probability to add student 3 to x is 2/14 because only 2 spots are left in this group. This way P(A2) = 1- 2/14 = 12/14 which is clearly wrong! can you please correct my misunderstanding? thank you in advance. – sel Apr 28 '20 at 18:17
  • You state "...we randomly pick 2 grad students and put them in group x" - why do you want to do this? The question asks you to calculate the probability that the graduate students are in 4 different groups. The first graduate student is assigned to a group. Let this group be group 1. Then $P(A_1)$ = probability that graduate student 2 is in a group other than group 1. Similarly, $P(A_2|A_1)$ = probability that graduate student 3 is in a group without graduate student 1 and 2; $P(A_3|A_1\cap A_2)$ = probability that graduate student 4 is in a group without graduate student 1, 2, and 3. – Michael R Apr 29 '20 at 19:16
  • @MichaelR Can you also use the additively axiom to arrive at this? The probability that graduate student 2 is placed in any slot is 1/15 and there are 12 slots that are not in group X so the probability of the union of each of these events is the sum of each probability, namely 12/15. Congratulations on completing medical school! – Kenny Nov 05 '21 at 23:52
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Yeah, this question gave me a headache for a week but then it was easy once I realized how to look at the question. The first catch is that the probability is being calculated for each group. Do not confuse with a sequence of events. The book has skipped a step as well. let's assume the initial state. Four groups and each group have 4 location that is valid for a grad student to occupy. The probability that a group will have a grad student is $\frac{4}{16}$ and overall probability of grad student in any of the group can be calculated as $\frac{4}{16} + \frac{4}{16} +\frac{4}{16} +\frac{4}{16} =\frac{16}{16} = 1 $. So the probability of having a grad in any valid position is one. Now we have occupied the first group with a valid position whats the probability of the second grad to have a valid position

Group 1: X _ _ _ Group 2: _ _ _ _ Group 3: _ _ _ _ Group 4: _ _ _ _

we have in total of 15 vacant positions out of which 3 are in group 1 that we can not be used and 12 are valid positions for second grad and the probability is $\frac{12}{15}$. Now we assume that second grad occupied valid positions we have the following Group 1: X _ _ _ Group 2: X _ _ _ Group 3: _ _ _ _ Group 4: _ _ _ _

now for 3rd grad student to be in a valid position we have 8 locations and the total slots are 14 so the probability is $\frac{8}{14}$. After the 3rd valid allocation, the final situation is as below Group 1: X _ _ _ Group 2: X _ _ _ Group 3: X _ _ _ Group 4: _ _ _ _

Four valid poistion in group 4 out of 13 so the probability of that group is $\frac{4}{13}$. Now multiply all. $1.\frac{12}{15}.\frac{8}{14}.\frac{4}{13}$

sami ullah
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