1
00:00:00,000 --> 00:00:01,040
2
00:00:01,040 --> 00:00:04,710
In this problem, we're given an
urn with n balls in it, out
3
00:00:04,710 --> 00:00:08,510
of which m balls
are red balls.
4
00:00:08,510 --> 00:00:13,370
To visualize it, we can draw a
box that represents the set of
5
00:00:13,370 --> 00:00:14,990
all n balls.
6
00:00:14,990 --> 00:00:18,310
Somewhere in the middle or
somewhere else we have a cut,
7
00:00:18,310 --> 00:00:21,750
such that to the left we have
all the red balls (there are
8
00:00:21,750 --> 00:00:24,670
m), and non-red balls.
9
00:00:24,670 --> 00:00:27,100
Let's for now call
it black balls.
10
00:00:27,100 --> 00:00:30,250
That is n minus m.
11
00:00:30,250 --> 00:00:34,450
Now, from this box, we are to
draw k balls, and we'd like to
12
00:00:34,450 --> 00:00:37,900
know the probability that
i out of those k
13
00:00:37,900 --> 00:00:39,455
balls are red balls.
14
00:00:39,455 --> 00:00:42,690
15
00:00:42,690 --> 00:00:44,670
For the rest of the problem,
we'll refer to this
16
00:00:44,670 --> 00:00:50,120
probability as p-r, where r
stands for the red balls.
17
00:00:50,120 --> 00:00:54,180
So from this picture, we know
that we're going to draw a
18
00:00:54,180 --> 00:00:59,330
subset of the balls, such that
i of them are red, and the
19
00:00:59,330 --> 00:01:03,920
remaining k minus i are black.
20
00:01:03,920 --> 00:01:06,830
And we'll like to know what is
the probability that this
21
00:01:06,830 --> 00:01:09,020
event would occur.
22
00:01:09,020 --> 00:01:13,360
To start, we define our sample
space, omega, as the set of
23
00:01:13,360 --> 00:01:17,150
all ways to draw k balls
out of n balls.
24
00:01:17,150 --> 00:01:19,540
We found a simple counting
argument -- we know that size
25
00:01:19,540 --> 00:01:23,750
of our sample space has
n-choose-k, which is the total
26
00:01:23,750 --> 00:01:25,710
number of ways to draw k
balls out of n balls.
27
00:01:25,710 --> 00:01:28,300
28
00:01:28,300 --> 00:01:30,450
Next, we'd like to know how
many of those samples
29
00:01:30,450 --> 00:01:32,630
correspond to the event that
we're interested in.
30
00:01:32,630 --> 00:01:35,700
In particular, we would like to
know c, which is equal to
31
00:01:35,700 --> 00:01:38,670
the number of ways to get
i red balls after
32
00:01:38,670 --> 00:01:40,800
we draw the k balls.
33
00:01:40,800 --> 00:01:44,500
To do so, we'll break c into
a product of two numbers --
34
00:01:44,500 --> 00:01:46,660
let's call it a times b --
35
00:01:46,660 --> 00:01:52,680
where a is the total number of
ways to select i red balls out
36
00:01:52,680 --> 00:01:54,520
of m red balls.
37
00:01:54,520 --> 00:02:02,530
So the number of ways to get
i out of m red balls.
38
00:02:02,530 --> 00:02:05,820
Going back to the picture, this
corresponds to the total
39
00:02:05,820 --> 00:02:09,340
number of ways to
get these balls.
40
00:02:09,340 --> 00:02:15,000
And similarly, we define b as
the total number of ways to
41
00:02:15,000 --> 00:02:23,150
get the remaining k minus i
balls out of the set n minus m
42
00:02:23,150 --> 00:02:25,340
black balls.
43
00:02:25,340 --> 00:02:29,890
This corresponds to the total
number of ways to select the
44
00:02:29,890 --> 00:02:34,820
subset right here in the
right side of the box.
45
00:02:34,820 --> 00:02:38,080
Now as you can see, once we have
a and b, we multiply them
46
00:02:38,080 --> 00:02:41,600
together, and this yields
the total number of ways
47
00:02:41,600 --> 00:02:43,470
to get i red balls.
48
00:02:43,470 --> 00:02:48,335
To compute what these numbers
are, we see that a is equal to
49
00:02:48,335 --> 00:02:54,280
m-choose-i number of ways to
get i red balls, and b is n
50
00:02:54,280 --> 00:03:01,570
minus m, the total number of
black balls, choose k minus i,
51
00:03:01,570 --> 00:03:06,160
the balls that are not red
within those k balls.
52
00:03:06,160 --> 00:03:09,670
Now putting everything back, we
have p-r, the probability
53
00:03:09,670 --> 00:03:14,260
we set out to compute, is equal
to c, the size of the
54
00:03:14,260 --> 00:03:20,250
event, divided by the size of
the entire sample space.
55
00:03:20,250 --> 00:03:22,700
From the previous calculations,
we know that c
56
00:03:22,700 --> 00:03:28,410
is equal to a times b, which
is then equal to m-choose-i
57
00:03:28,410 --> 00:03:33,000
times (n minus m)-choose-(k
minus i).
58
00:03:33,000 --> 00:03:38,460
And on the denominator, we have
the entire sample space
59
00:03:38,460 --> 00:03:41,110
is a size n-choose-k.
60
00:03:41,110 --> 00:03:44,810
And that completes
our derivation.
61
00:03:44,810 --> 00:03:48,080
Now let's look at a numerical
example of this problem.
62
00:03:48,080 --> 00:03:51,630
Here, let's say we have
a deck of 52 cards.
63
00:03:51,630 --> 00:03:57,290
And we draw a box with n equals
52, out of which we
64
00:03:57,290 --> 00:04:00,400
know that there are 4 aces.
65
00:04:00,400 --> 00:04:04,660
So we'll call these the left
side of the box, which is we
66
00:04:04,660 --> 00:04:09,130
have m equals 4 aces.
67
00:04:09,130 --> 00:04:12,640
Now if we were to draw
seven cards--
68
00:04:12,640 --> 00:04:15,550
call it k equal to 7--
69
00:04:15,550 --> 00:04:18,620
and we'd like to know what is
the probability that out of
70
00:04:18,620 --> 00:04:21,340
the 7 cards, we have 3 aces.
71
00:04:21,340 --> 00:04:24,610
72
00:04:24,610 --> 00:04:29,420
Using the notation we did
earlier, if we were to draw a
73
00:04:29,420 --> 00:04:33,000
circle representing the seven
cards, we want to know what is
74
00:04:33,000 --> 00:04:36,550
the probability that we have 3
aces in the left side of the
75
00:04:36,550 --> 00:04:41,650
box and 4 non-aces for the
remainder of the deck.
76
00:04:41,650 --> 00:04:46,570
In particular, we'll
call i equal to 3.
77
00:04:46,570 --> 00:04:49,990
So by this point, we've cast the
problem of drawing cards
78
00:04:49,990 --> 00:04:53,810
from the deck in the same way
as we did earlier of drawing
79
00:04:53,810 --> 00:04:55,210
balls from an urn.
80
00:04:55,210 --> 00:04:59,970
And from the expression right
here, which we computed
81
00:04:59,970 --> 00:05:03,630
earlier, we can readily compute
the probability of
82
00:05:03,630 --> 00:05:06,280
having 3 aces.
83
00:05:06,280 --> 00:05:10,360
In particular, we just have to
substitute into the expression
84
00:05:10,360 --> 00:05:16,370
right here the value of m equal
to 4, n equal to 52, k
85
00:05:16,370 --> 00:05:19,980
equal to 7, finally,
i equal to 3.
86
00:05:19,980 --> 00:05:29,300
So we have 4-choose-3 times n
minus m, in this case would be
87
00:05:29,300 --> 00:05:36,260
48, choose k minus i, will be 4,
and on the denominator, we
88
00:05:36,260 --> 00:05:41,750
have 52 total number of cards,
choosing 7 cards.
89
00:05:41,750 --> 00:05:42,810
That gives us [the]
90
00:05:42,810 --> 00:05:44,440
numerical answer [for]
91
00:05:44,440 --> 00:05:47,890
the probability of getting 3
aces when we draw 7 cards.
92
00:05:47,890 --> 00:05:49,140