Student council question

Jaspworld

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Apr 17, 2007
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Ques.) The student council consists of two grade 9 students, three grade 10 students, four grade 11 students, and five grade 12 students. A committee of 4 is formed by placing all 14 names in a hat and drawing 4 names. What is the probability that the members of the committee are all from different grades.

My solution:

(2/14)*(3/13)*(4/12)*(5/11)*P(4,4)=120/2001

That's the right answer but why do you have to use P(4,4) at the end.
 
Hello, Jaspworld!

The student council consists of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors.
A committee of four is formed by placing all 14 names in a hat and drawing four names.
What is the probability that the members of the committee are all from different grades?

My solution: \(\displaystyle \:\frac{2}{14}\cdot\frac{3}{13}\cdot\frac{4}{12}\cdot\frac{5}{11}\cdot P(4,4) \;=\;\frac{120}{1001}\)

That's the right answer but why do you have to use P(4,4) at the end?

You said it's your solution . . . Why did you put it there?


Here's another approach . . .

There are: \(\displaystyle \:{14\choose4}\:=\:\L1001\) possible comittees.

There are 2 choices for a freshman, 3 choices for a sophomore,
. . 4 choices for a junior, and 5 choices for a senior.
Hence, there are: \(\displaystyle \,2\,\times\,3\,\times\,4\,\times5\:=\:\L120\) ways to choose one from each grade.

Therefore, the probabiity is: \(\displaystyle \L\:\frac{120}{1001}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Okay, to answer your question . . .

Your original product of fractions puts the choices in an order.

You started with \(\displaystyle \frac{2}{14}\)
This is the probability that the first name drawn is a freshman.

Then you had \(\displaystyle \frac{3}{13}\), the probability that the second name drawn is a sophomore.

\(\displaystyle \frac{4}{12}\) is the probability that the third name is a junior.
\(\displaystyle \frac{5}{11}\) is the probability that the fourth name is a senior.


Then: \(\displaystyle \:\frac{2}{14}\,\times\,\frac{3}{13}\,\times\,\frac{4}{12}\,\times\,\frac{5}{11}\:=\:\frac{5}{1001}\) is the probability that the names drawn are
. . a freshman, a sophomore, a junior, and a senior . . . in that order.


Since the order of names is irrelevant, and there are \(\displaystyle P(4,4)\,=\,24\) orders,

. . the probability is: \(\displaystyle \:24\,\times\,\frac{5}{1001}\;=\;\frac{120}{1001}\)

 
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