There are STUDENTS students in a class: BOYS boys and
GIRLS girls.
If the teacher picks a group of GROUP at random, what is the probability that
everyone in the group is a boy?
If the teacher picks a group of GROUP at random, what is the probability that
everyone in the group is a girl?
One way to solve this problem is to figure out how many different groups there are of only boys, then divide this by the total number of groups you could choose. Since every group is chosen with equal probability, this will be the probability that a group of all boys is chosen.
One way to solve this problem is to figure out how many different groups there are of only girls, then divide this by the total number of groups you could choose. Since every group is chosen with equal probability, this will be the probability that a group of all girls is chosen.
We know two ways to count the number of groups we can choose: we use permutations if order matters, and combinations if it doesn't. Does the order the students are picked matter in this case?
It doesn't matter if we pick John then
Ben or Ben then
John,
so order must not matter. So, the number of ways
to pick a group of GROUP students out of STUDENTS is
\dfrac{STUDENTS!}{(STUDENTS-GROUP)!GROUP!} =
\binom{STUDENTS}{GROUP}.
[Show me why]
It doesn't matter if we pick Julia then
Beatrice or Beatrice then
Julia,
so order must not matter. So, the number of ways
to pick a group of GROUP students out of STUDENTS is
\dfrac{STUDENTS!}{(STUDENTS-GROUP)!GROUP!} =
\binom{STUDENTS}{GROUP}.
[Show me why]
Remember that the
STUDENTS! \; and (STUDENTS-GROUP)! \; terms come from when we
fill up the group, making STUDENTS
choices for the first slot, then STUDENTS-1 choices for the
second, and so on. In this way, we end up making
_.map(_.range(GROUP), function(l){ return (STUDENTS-l);}).join("\\cdot")
= \dfrac{STUDENTS!}{(STUDENTS-GROUP)!} \;.
The GROUP! \; term comes from the number of times we've counted
a group as different because we chose the students in a different order.
There are GROUP! \;
ways to order a group of GROUP, so for every group, we've overcounted exactly
that many times.
We can use the same logic to count the number of groups that only have boys.
We can use the same logic to count the number of groups that only have girls.
Specifically, the number of ways to pick a group of GROUP students out of
NUM_B is
\dfrac{NUM_B!}{(NUM_B-GROUP)!GROUP!} =
\binom{NUM_B}{GROUP}.
So, the probability that the teacher picks a group of all boys is the number of groups with only boys divided by the number of total groups the teacher could pick.
So, the probability that the teacher picks a group of all girls is the number of groups with only girls divided by the number of total groups the teacher could pick.
This is \displaystyle \frac{\frac{NUM_B!}{(NUM_B-GROUP)!\cancel{GROUP!}}}
{\frac{STUDENTS!}{(STUDENTS-GROUP)!\cancel{GROUP!}}} =
\frac{\frac{NUM_B!}{NUM_B-GROUP!}}{\frac{STUDENTS!}{STUDENTS-GROUP!}}
We can re-arrange the terms to make simplification easier
\left(\dfrac{NUM_B!}{NUM_B-GROUP!}\right)
\left(\dfrac{STUDENTS-GROUP!}{STUDENTS!}\right) =
\left(\dfrac{NUM_B!}{STUDENTS!}\right)
\left(\dfrac{STUDENTS-GROUP!}{NUM_B-GROUP!}\right)
Simplifying, we get
\left(\dfrac{\cancel{NUM_B!}}{_.map(_.range(STUDENTS-NUM_B), function(l){ return (STUDENTS-l); }).join("\\cdot")
\cdot \cancel{NUM_B!}}\right)
\left(\dfrac{_.map(_.range(STUDENTS-NUM_B), function(l){ return (STUDENTS-GROUP-l); }).join("\\cdot")
\cdot \cancel{(NUM_B-GROUP)!}}{\cancel{NUM_B-GROUP!}}\right) =
\left(\dfrac{1}{factorial(STUDENTS)/factorial(NUM_B)}\right)
\left(factorial(STUDENTS-GROUP)/factorial(NUM_B-GROUP)\right) =
\dfrac{PRETTY_NUM}{PRETTY_DEM}
If you flip a fair coin COINS times, what is the probability that you will get exactly
NUM NAME?
PRETTY_NUMER/PRETTY_DENOM
One way to solve this problem is to figure out how many ways you can get exactly NUM NAME, then
divide this by the total number of outcomes you could have gotten. Since every outcome has equal probability, this will be the
probability that you will get exactly
NUM NAME.
How many outcomes are there where you get exactly NUM NAME? Try thinking of each outcome as
a COINS-letter word, where the first letter is "H" if the first coin toss was heads and "T"
if it was tails, and so on.
So, the number of outcomes with exactly NUM NAME
is the same as the number of these words which have
NUM H's and COINS-NUM T's.
So, the number of outcomes with exactly NUM NAME
is the same as the number of these words which have
NUM T's and COINS-NUM H's.
How many of these are there? If we treat all the letters as unique,
we'll find that there are COINS! different arrangements, overcounting NUM!
times for every time we only switch the H's around, and COINS-NUM!
times for every time we only switch the T's around.
[Show me why]
How many of these are there? If we treat all the letters as unique,
we'll find that there are COINS! different arrangements, overcounting NUM!
times for every time we only switch the T's around, and COINS-NUM!
times for every time we only switch the H's around.
[Show me why]
5\cdot4\cdot3\cdot2\cdot1 = 5! = 120 \; different re-arrangements. But, this treats all the letters
as unique, when 3! of these multi-
colored arrangements for every normal one, so that means dividing
our first guess of 5! by 3!. By the exact same logic, we need to divide by 2! to
avoid overcounting every permutation where we just move the heads around. So, the number of re-arrangements is
\dfrac{5!}{3!2!} = \binom{5}{3}.
So, there are \dfrac{COINS!}{NUM!COINS-NUM!} = NUM_RIGHT different
outcomes where you get exactly NUM NAME.
[How many total outcomes are there?]
Altogether, there are 2^{COINS} = NUM_ALL total possible outcomes.
So, the probability that you will get exactly NUM NAME is
\dfrac{NUM_RIGHT}{NUM_ALL} = \dfrac{PRETTY_NUMER}{PRETTY_DENOM}.
So, the probability that you will get exactly NUM NAME is
\dfrac{NUM_RIGHT}{NUM_ALL}.