WEBVTT
1
00:00:00.910 --> 00:00:02.690 A:middle L:90%
all right. In a previous problem, we were
2
00:00:02.690 --> 00:00:05.129 A:middle L:90%
given a bunch of probabilities that a certain firm will
3
00:00:05.129 --> 00:00:08.289 A:middle L:90%
get awarded certain projects and we're supposed to now find
4
00:00:08.289 --> 00:00:11.240 A:middle L:90%
some conditional probabilities as well as interpret them in the
5
00:00:11.240 --> 00:00:15.679 A:middle L:90%
context of the problem. So let's start with party
6
00:00:16.129 --> 00:00:18.250 A:middle L:90%
. The probability of a two given a one.
7
00:00:18.399 --> 00:00:20.519 A:middle L:90%
And for this we're gonna use our definition of conditional
8
00:00:20.519 --> 00:00:23.199 A:middle L:90%
probability. So that's gonna be probability of a two
9
00:00:23.429 --> 00:00:26.960 A:middle L:90%
intersect a one all over the probability of a one
10
00:00:29.140 --> 00:00:31.539 A:middle L:90%
were given both of these values. The probability of
11
00:00:31.550 --> 00:00:35.799 A:middle L:90%
a two intersect A one is 0.11 The probability of
12
00:00:35.799 --> 00:00:39.000 A:middle L:90%
a one is there a 10.22? This reduces down
13
00:00:39.000 --> 00:00:43.170 A:middle L:90%
to 1/2 or 0.5. What this means in the
14
00:00:43.170 --> 00:00:45.630 A:middle L:90%
context of the problem is that there is a 50%
15
00:00:45.630 --> 00:00:48.149 A:middle L:90%
chance that, given the firm is awarded Project number
16
00:00:48.149 --> 00:00:51.070 A:middle L:90%
one, it will also be awarded Project number two
17
00:00:53.490 --> 00:00:55.520 A:middle L:90%
. All right, next problem. We have the
18
00:00:55.520 --> 00:00:59.250 A:middle L:90%
probability of a two intersect a three given a one
19
00:00:59.939 --> 00:01:00.850 A:middle L:90%
. Once the debt, once again, we're gonna
20
00:01:00.850 --> 00:01:07.760 A:middle L:90%
use our definition of conditional probability eight to intercept a
21
00:01:07.760 --> 00:01:14.480 A:middle L:90%
three intersect a one all over the probability of a
22
00:01:14.480 --> 00:01:19.379 A:middle L:90%
one. All right. Since Intersection is both communities
23
00:01:19.400 --> 00:01:23.349 A:middle L:90%
and associative, this is the same thing as the
24
00:01:23.349 --> 00:01:26.939 A:middle L:90%
probability of a one intersect h, you intersect a
25
00:01:26.939 --> 00:01:34.920 A:middle L:90%
three, which turns out to be 0.1 over 0.22
26
00:01:34.920 --> 00:01:37.819 A:middle L:90%
for a probability of a one. And this is
27
00:01:37.829 --> 00:01:42.980 A:middle L:90%
0.0 455 Now, what this means is that,
28
00:01:42.980 --> 00:01:46.950 A:middle L:90%
given the given that the firm is awarded project number
29
00:01:46.950 --> 00:01:49.680 A:middle L:90%
one, there is a 4.55% chance that the same
30
00:01:49.680 --> 00:01:53.599 A:middle L:90%
form will be awarded both projects too. And three
31
00:01:53.609 --> 00:01:56.959 A:middle L:90%
, Remember that intersection generally means. And when we're
32
00:01:56.959 --> 00:02:00.590 A:middle L:90%
interpreting a problem like this, now we have probability
33
00:02:00.590 --> 00:02:05.370 A:middle L:90%
of 18 Union A three given a one. Once
34
00:02:05.370 --> 00:02:07.960 A:middle L:90%
again, we can use our definition of conditional probability
35
00:02:15.919 --> 00:02:17.509 A:middle L:90%
. All right, now, if we look at
36
00:02:17.509 --> 00:02:21.509 A:middle L:90%
this, we look at this top and we see
37
00:02:21.580 --> 00:02:23.009 A:middle L:90%
, huh? Intersection distributes so we can expand that
38
00:02:23.009 --> 00:02:38.169 A:middle L:90%
out. Union A three, a one. There
39
00:02:38.169 --> 00:02:42.090 A:middle L:90%
we go. All right. Now we can use
40
00:02:42.090 --> 00:02:45.620 A:middle L:90%
our descent rule for probability to expand the numerator two
41
00:02:45.620 --> 00:02:53.990 A:middle L:90%
probability of a to intersect a one plus probability of
42
00:02:53.000 --> 00:02:59.650 A:middle L:90%
a three intersect. A one minus probability of would
43
00:02:59.650 --> 00:03:01.689 A:middle L:90%
be the intersection of both of these. But that's
44
00:03:01.689 --> 00:03:05.979 A:middle L:90%
just a one. Intersect H two intersect a three
45
00:03:07.219 --> 00:03:13.229 A:middle L:90%
because the A one is redundant all over the probability
46
00:03:13.229 --> 00:03:17.620 A:middle L:90%
of a one. So that's gonna be 0.11 plus
47
00:03:17.620 --> 00:03:27.340 A:middle L:90%
0.5 minus 0.1 all over 0.22 just substituting it.
48
00:03:27.340 --> 00:03:30.539 A:middle L:90%
And that ends up being 0.682 In the context of
49
00:03:30.539 --> 00:03:34.259 A:middle L:90%
this problem, what this means is that given that
50
00:03:34.259 --> 00:03:37.139 A:middle L:90%
the firm is awarded project number one, there is
51
00:03:37.139 --> 00:03:42.159 A:middle L:90%
a 68.2% chance that'll either be awarded Project One or
52
00:03:42.389 --> 00:03:44.830 A:middle L:90%
project number three. Sorry, Project number two or
53
00:03:44.830 --> 00:03:49.050 A:middle L:90%
project number three. Keep in mind a union generally
54
00:03:49.050 --> 00:03:52.560 A:middle L:90%
means or all right, Finally, we have the
55
00:03:52.789 --> 00:03:55.759 A:middle L:90%
business. The probability of a one intersect A to
56
00:03:55.770 --> 00:03:59.659 A:middle L:90%
intercept a three, given a one union A to
57
00:03:59.659 --> 00:04:05.393 A:middle L:90%
union A three once again using our definition of conditional
58
00:04:05.393 --> 00:04:12.782 A:middle L:90%
probability, it's a one intersect A to intercept a
59
00:04:12.782 --> 00:04:17.793 A:middle L:90%
three intersect a one union, a choo union,
60
00:04:17.802 --> 00:04:25.362 A:middle L:90%
a three all over probability of a one union A
61
00:04:25.362 --> 00:04:29.702 A:middle L:90%
to union a three. Now, in order,
62
00:04:29.702 --> 00:04:30.043 A:middle L:90%
some fly this top. I'm going to use a
63
00:04:30.043 --> 00:04:32.752 A:middle L:90%
Venn diagram is a visual, so let's call a
64
00:04:32.752 --> 00:04:36.062 A:middle L:90%
one the green. A two Sorry, a one
65
00:04:36.062 --> 00:04:39.083 A:middle L:90%
Laurent H of the green in a three, the
66
00:04:39.083 --> 00:04:41.942 A:middle L:90%
blue. And if we make a Venn diagram like
67
00:04:41.942 --> 00:04:53.483 A:middle L:90%
so where that's these colors correspond to each of these
68
00:04:53.483 --> 00:04:56.752 A:middle L:90%
ays. Then we know that a one intersect A
69
00:04:56.752 --> 00:05:00.233 A:middle L:90%
. To intercept a three is the center section here
70
00:05:01.153 --> 00:05:03.942 A:middle L:90%
and that a one union A to union A three
71
00:05:04.242 --> 00:05:10.492 A:middle L:90%
is the entirety of what's inside these circles. Now
72
00:05:10.802 --> 00:05:12.142 A:middle L:90%
, in order to find the intersection, we need
73
00:05:12.142 --> 00:05:14.372 A:middle L:90%
to figure out what's common between the two. Well
74
00:05:14.562 --> 00:05:15.913 A:middle L:90%
, that's just a one intersect A to intercept a
75
00:05:15.913 --> 00:05:19.872 A:middle L:90%
three. So this top, it's just the probability
76
00:05:19.872 --> 00:05:23.663 A:middle L:90%
of a one intersect a two, and it's like
77
00:05:23.673 --> 00:05:27.452 A:middle L:90%
a three. Now we have to use our addition
78
00:05:27.452 --> 00:05:30.182 A:middle L:90%
rule for three variables on the bottom Here, probability
79
00:05:30.182 --> 00:05:34.353 A:middle L:90%
of a one intersect. Sorry. Plus the probability
80
00:05:34.353 --> 00:05:40.622 A:middle L:90%
of a two lost the probability of a three minus
81
00:05:40.622 --> 00:05:45.733 A:middle L:90%
. The probability of a one intersect a two minus
82
00:05:45.733 --> 00:05:48.572 A:middle L:90%
The probability of a one intersect a three minus The
83
00:05:48.572 --> 00:05:55.853 A:middle L:90%
probability of a two intersect a three plus the probability
84
00:05:56.333 --> 00:06:00.711 A:middle L:90%
of a one intersect h you intersect a three by
85
00:06:00.711 --> 00:06:02.682 A:middle L:90%
the way. Yes. I am aware this was
86
00:06:02.682 --> 00:06:05.502 A:middle L:90%
a require part of the problem where these probabilities come
87
00:06:05.502 --> 00:06:08.221 A:middle L:90%
from. But that's just a jog. Your memory
88
00:06:08.221 --> 00:06:10.942 A:middle L:90%
, in case you don't remember that we did that
89
00:06:10.942 --> 00:06:12.752 A:middle L:90%
earlier or if he just wanted to computer by hand
90
00:06:12.752 --> 00:06:16.882 A:middle L:90%
again. All right. This top part is 0.1
91
00:06:18.112 --> 00:06:24.281 A:middle L:90%
and this bottom part ends up being 0.22 plus 0.25
92
00:06:24.291 --> 00:06:31.281 A:middle L:90%
plus 0.28 minus 0.11 minus 0.5 minus zero points.
93
00:06:32.072 --> 00:06:34.211 A:middle L:90%
Said then that doesn't work, right? It should
94
00:06:34.211 --> 00:06:38.851 A:middle L:90%
be 07 shouldn't it? All right. Yeah,
95
00:06:38.862 --> 00:06:41.512 A:middle L:90%
that should be 07 My fault. Just looking back
96
00:06:41.512 --> 00:06:43.182 A:middle L:90%
at the problem, I just realized that looked way
97
00:06:43.182 --> 00:06:46.141 A:middle L:90%
too big. Plus the probability of zero Windsor one
98
00:06:46.302 --> 00:06:49.382 A:middle L:90%
that ends up being 0.53 If you divide that out
99
00:06:49.382 --> 00:06:56.541 A:middle L:90%
, that become 0.189 Now, what does this mean
100
00:06:56.541 --> 00:06:58.341 A:middle L:90%
in the context of the problem? Well, that
101
00:06:58.341 --> 00:07:01.692 A:middle L:90%
means that given that this project this firm got awarded
102
00:07:01.732 --> 00:07:04.541 A:middle L:90%
either projects 12 or three, that there is a
103
00:07:04.541 --> 00:07:10.701 A:middle L:90%
1.89% chance that it got awarded all three projects projects
104
00:07:10.701 --> 00:07:13.841 A:middle L:90%
one and two and three. And there you have
105
00:07:13.841 --> A:middle L:90%
it