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So here's the problem:

There are $100$ students who want to sign up for the class Introduction to Acting. There are three class sections for Introduction to Acting, each of which will fit exactly $20$ students. The $100$ students including Dendi and Nico, are put in a lottery, and $60$ of them are randomly selected to fill up the classes.

What is the probability that Dendi and Nico end up getting into the same section for the class? The answer for this problem is $\frac{19}{165}$, and I am tasked with finding the solution.

I don't know how to approach this problem. I'm guessing that we must calculate the number of ways to get 60 students from a group of 100, so $^{100}P _{60}.$ I guess we need to get the result of that, let's assume is $a$, and do $^{60}P_{20}$? I'm not sure. Any help would be appreciated.

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    $\begingroup$ So, there is a $\frac {60}{100}$ chance that $D$ is picked for a class. There are then $19$ other seats in the same class...so the probability that $S$ is also picked for that class is... $\endgroup$
    – lulu
    Apr 5, 2021 at 15:06
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    $\begingroup$ No. There are $19$ open slots and $99$ people to populate them with, so the probability that $S$ gets one of the slots is... $\endgroup$
    – lulu
    Apr 5, 2021 at 15:56
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    $\begingroup$ Well, you need both events to happen, so now you just multiply those two fractions. $\endgroup$
    – lulu
    Apr 5, 2021 at 16:03
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    $\begingroup$ Perhaps you would find it clearer if I phrased it this way: The probability that they are both assigned class $\#1$ is $\frac {20}{100}\times \frac {19}{99}$, yes? Same for the other two classes. So now just multiply by $3$. $\endgroup$
    – lulu
    Apr 5, 2021 at 16:28
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    $\begingroup$ @lulu: I would suggest that you give the solution of $\frac{60}{100}\frac{19}{99}$ as a coherent answer; here it lies scattered in bits and pieces. $\endgroup$ Apr 5, 2021 at 17:38

3 Answers 3

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The same problem can be approached in various ways, and obviously the simpler the computation, the better

An instructive illustration worth seeing, of the variety of ways the same question can be approached is here on this forum, where alternative approaches were specifically asked for.

One idea used was to place the objects of interest one by one without bothering about what happens to the others.

Using the same approach here, the reqd. probability is simply $\frac{60}{100}\frac{19}{99}$

The first fraction is the Pr for one friend to get a place in some class, and the second fraction places the other friend in the same class

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Lets call the classes as $A,B,C$ and lets say that both of them are in the class $A$.

Then , we should chose $18$ student from the out of $98$ students for the rest of class $A$ . Moreover , select $20$ students for $B ,C$ such that $C(80,20) \times C(60,20)$

As a result we have to make selection like $ C(98,18)\times C(80,20) \times C(60,20)$ when the students in $A$ ,but there are $3$ classes so , it will be $3 \times C(98,18)\times C(80,20) \times C(60,20)$.

Then , the result is $\frac{3 \times C(98,18)\times C(80,20) \times C(60,20)}{C(100,20)\times C(80,20) \times C(60,20)}$.

Lets cancel out this huge expression , as you see , there are common terms in both of numerator and denominator such as $C(80,20) \times C(60,20)$.

The neat form will be like $\frac{3 \times C(98,18)}{C(100,20)}=\frac{3 \times 98! \times 80! \times 20!}{80! \times 18! \times 100!}=19/165$

REMINDER: When you select something , use combination instead of permutation unless the order mentioned

NOTE: This way is a little cumbersome , i wrote it to show another perspective. For practical way , @lulu's comment is elegant

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    $\begingroup$ You are making this way too complicated. Look at lulu's comment on the question. $\endgroup$
    – saulspatz
    Apr 5, 2021 at 15:38
  • $\begingroup$ Have you tried computing this number? $\endgroup$
    – lulu
    Apr 5, 2021 at 16:00
  • $\begingroup$ @lulu hımm, it is interesting. Thanks for this warning. However , it seems true . where did i make wrong $\endgroup$ Apr 5, 2021 at 16:03
  • $\begingroup$ Your denominator ought to be $\binom {100}{20}\times \binom {80}{20}\times \binom {60}{20}$. And then, of course, the expression simplifies enormously. $\endgroup$
    – lulu
    Apr 5, 2021 at 16:10
  • $\begingroup$ @lulu oh yeah , how did i miss it ! .Thank you .. $\endgroup$ Apr 5, 2021 at 16:12
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This query shows that intuition can be a powerful weapon in problems involving Probability, Combinatorics, or Statistics.

From saulspatz

You are making this way too complicated. Look at lulu's comment on the question.

From lulu

Perhaps you would find it clearer if I phrased it this way: The probability that they are both assigned class #1 is
$\displaystyle \frac{20}{100} \times \frac{19}{99}$?
Same for the other two classes. So now just multiply by 3.

$\displaystyle 3 \times \frac{20}{100} \times \frac{19}{99} $

$\displaystyle =~ 3 \times \frac{1}{5} \times \frac{19}{99} $

$\displaystyle =~ 3 \times \frac{19}{99} \times \frac{1}{5} $

$\displaystyle =~ \frac{19}{33} \times \frac{1}{5} $

$\displaystyle =~ \frac{19}{165}. $

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