+0  
 
0
108
4
avatar

In a class of 28 students, the teacher selects four people at random to participate in a geography contest. What is the probability that this group of four students includes at least two of the top three geography students in the class? Express your answer as a common fraction.

 

My solution was 3C2 * 26C2  and to divide by the total number of groups but its incorrect. Why? Where am I overcounting? I already know about the correct/other solution where you seperate the cases into either all top 3 or only 2 top 3 and add the cases. I just want to know specifically why this method was wrong.

 Feb 9, 2022
 #1
avatar+2331 
0

At the risk of sounding like a Troll, (which I would never, ever do because it’s so rude), the specific reasons your method is wrong are: ...

 Feb 9, 2022
 #2
avatar+2331 
0

You didn’t set up any equations, so you don’t know what the hell is going on. You plopped two (2) combinations for multiplication with a comment about dividing.

 

In your plop, you removed the three (3) top students for selection (3C2), but then included one of the top students in the combination for non-top students. 

 

-------

To solve this:

Note that the “At Least” condition creates a mathematical “OR,” requiring the creation of two (2) sets: The first set has two of the top students; the second has all three top students. These sets are added giving the total number of sets with two (2) OR three (3) top students

 

 

\(\dbinom{3}{2}*\dbinom{25}{2} = 900 \hspace {1em} | \hspace {2em} \text {Number of sets having two of the three top students.}\\ \dbinom{3}{3}*\dbinom{25}{1} = 25 \hspace {1em} | \hspace {2em} \text {Number of sets having all three of the top students.}\\ \text { }\\ 900 + 25 =925 \\ \text {Divide this by the total number of sets of four (4) from (28): } \binom{28}{4} = 20475 \\ \LARGE \rho \normalsize (r\geq 2)  = \dfrac{925}{20475} \rightarrow \dfrac{37}{819}\)

 

 

GA

--. .-

GingerAle  Feb 9, 2022
edited by GingerAle  Feb 9, 2022
edited by GingerAle  Feb 11, 2022
 #3
avatar
0

"In your plop, you removed the three (3) top students for selection (3C2), but then included one of the top students in the combination for non-top students. "

 

I included a top student (student that wasn't chosen in the 3C2) in the combination for non top students because I also wanted to include the cases where all top 3 and another student are on the team. Knowing this, could you elaborate on my mistake?

Guest Feb 10, 2022
 #4
avatar+2331 
0

Moving the unselected top student into the sample space of the non-top students results in over counting. Specifically, the sample space increases to 26 (nCr(26, 2)) and the top student’s selection (as one of the two selected) occurs 25 times out of the 325. This is (1/13) of the total selections and (12/13) of the selections should not be counted as sets containing the three (3) top students. Note that while 25 matches the second binomial product for the “OR” condition, multiplying this by your first binomial results in 75.

 

Points to note:

When isolating sample space elements for a sub-selection, the values of (n) in each binomial will sum to the total sample space (n). Note the (3) and (25) in \(\binom{3}{2}*\binom{25}{2}\;and\; \binom{3}{3}*\binom{25}{1}\)  sum to (28), the total sample space.  Note also the (k) selection in each binomial sums to four (4) the total selection (k).

 

Isolated sample space elements for each sub-selection are added or subtracted, depending on the case. The result(s) are then divided by the selections from total sample space.

 

 

GA

--. .-

GingerAle  Feb 11, 2022

19 Online Users

avatar