Uniform Probability Distributions (movie 5.3)

# Uniform Probability Distributions (movie 5.3)

This video is devoted to uniform probability
distributions. This was a concept introduced in the last
video. Remember a uniform sample space is one where
all the elements have equal probabilities. This occurs so commonly in the real world
so that we’re focusing on this and seeing what sort of special techniques are available
whenever all the probabilities are the same. Let’s start by thinking about a club electing
a president and vice president. The club has 14 members which includes 6 men
and 8 women. Assuming that all members are equally likely
to be elected, what is the probability that both officers will be women? Here’s a way to think about that. Let’s just focus on which 2 people are elected
without thinking about who is the president and who’s the vice president. The question only asks what’s the chances
the 2 officers will be women, so we really don’t have to think about who is president
and who is vice president in order to answer the question. We have 14 people and 2 of them are going
to be elected. And from our basic counting techniques we
know using the combinations formula that the number of different ways of choosing 2 people
from 14 can be easily obtained by the combination formula, turns out to be 91. So there are 91 different ways that 2 people
could be selected from the 14 as the officers. This is again not distinguishing between who
is president and who is vice president, just talking about selecting 2 from the 14. If you were going to pick 2 officers from
the 8 women you would of course be selecting the 2 people from the set of 8 women. And the combinations formula gives us 28 ways
of doing that. So what we know now is that there are 91 possible
outcomes and 28 of those 91 outcomes would be cases where both officers are women. The answer to this question is 28 over 91. And the reason is because we’re dealing with
91 equally likely outcomes. So the probability of each one of these 91
cases would be 1 over 91, 28 of those cases, so their probabilities would add up to 28
over 91. Now let’s do an alternate solution in which
we do keep track of who’s elected to which office. So we’re going to be thinking permutations
instead of combinations in this solution. We have the 14 members, we’re going to select
1 person as president and 1 person as vice president. Let’s first think about who’s selected for
president. 14 choices for president, and then after one
of those is selected, that leaves 13 choices for vice president. 14 choices for president times 13 choices
for vice president is permutations of 14 club members, choosing 2 of them. An ordered list, because the first one is
going to be president the second vice president. So keeping track of who fills which office,
we get 182 as the number of possibilities for who could be president and secondly vice
president. If we do it in such a way that both are women,
then we have 8 choices for which woman serves as president as well as 7 choices which serves
as vi! ce president giving us 56 ways of filling the 2 offices consecutively with 2 women. 182 equally likely outcomes, 56 of those outcomes
being cases where both officers are women. 56 over 182 is the same as 28 over 91 which
is the same answer we obtained before. Now this should be reassuring to you, because
we arrive at the correct answer, no matter whether we keep track of who fills each office
or whether we ignore who fills each office. In other words, we can treat this particular
problem either by just considering which 2 people are elected or else by specifically
concerning ourselves with who’s president and who’s vice president. The problem doesn’t require us to keep track
of that more detailed information but we arrive at the correct answer either way. In both approaches in effect what we did was
to count the total number of possible outcomes and put that in the denominator, and count
the number of outcomes that make up the event that we’re studying and put that in the numerator. The probability of that event occurring is
just the number of ways that that thing can happen divided by the total number of possible
outcomes. And the idea is as simple as in this example. If you’re dealing from a 52 card deck then
the chances of it being a heart is 13 over 52, one fourth because there are 13 cards
in the 52 card deck. The probability is one fourth because one
fourth of the cards are hearts. Charlie has a bunch of socks that are mixed
up in his drawer where he keeps his socks. He has 4 pairs of white socks, 2 pairs of
black socks, and 3 pairs of blue socks. With these socks all scrambled up, he gets
up in the morning in the dark, reaches in the drawer and pulls out 2 socks. What’s the probability the socks will be of
the same color? Well how many socks does he have? 4 plus 2
plus 3 is 9 pairs of socks, so that means 18 socks. And he’s picking 2 socks because he has 2
feet. So he’s picking 2 socks from the 18 socks
in the drawer. And the number of ways of picking 2 socks
from 18 is given by the combinations formula. 18 times 17 divided by 2 is 153. So that’s the total number of possible outcomes
when he reaches in his drawer and pulls out 2 socks. Now we’re interested in the probability the
socks will be the same color. There are 3 ways that can happen. It can happen if he gets 2 white socks, it
can happen if he gets 2 black socks. Or it will happen if he gets 2 blue socks. To get 2 white socks, the 2 socks he’s picking
have to come from the 8 white socks that are in the drawer and that can happen in 28 ways
according to the combinations formula. There are only 4 black socks in the drawer
so to get 2 black socks, the 2 socks he’s picking have to come from those 4. The combinations formula gives us 6 ways that
can happen. And with 6 blue socks we get 15 ways that
he can get a pair of blue socks. If we total all these up, 28 plus 6 plus 15
we have a total of 49 different outcomes in which his socks match. That’s 49 out of the 153 total number of possible
outcomes. So that gives us a probability of 49 over
153 for our answer, so that’s just a little bit less than a third. A l! ittle less than a third is the chances
that Charlie gets a matching pair of socks. Eric and Edith have applied for jobs at Sadlacks. There are 6 job openings available, 25 people
have applied. First question is, what’s the probability
that Edith gets a job? And then secondly what’s the chances they both get jobs. And our instructions here are to assume that
the 6 people who get jobs are to be randomly chosen from among the 25 applicants. For part A what’s the probability that Edith
gets the job. We’re giving 6 jobs out among 25 people so
we start by counting all the possible ways that you can choose which 6 of the 25 people
gets jobs. That’s a job for the combinations formula. Plug that in combinations of choosing 6 things
from 25. Combinations formula cranks out this many
ways to pick the 6. Now of that total number, very large number
of possible outcomes, how many of those cases would be a case where Edith gets a job? If
Edith gets one of the jobs then that means there will be 5 other people who get jobs,
right? There are 6 jobs altogether so if Edith gets one of the jobs, that means 5 other people
will get jobs from among the other 24 people. 1 of the 25 is Edith. So there will be 5 other jobs to be given
out among the other 24 people besides Edith. So the number of ways that Edith can get a
job is just a question of how many ways you pick which other 5 people from the other!
24 people get jobs. Plug that into the combinations formula, determine
how many ways there are of choosing that. You come out with 42 thousand and some. So it would be the quotient of those 2 numbers
that reduces to 6 over 25 and it’s worth doing the arithmetic to see why it reduces to that
rather than working with these very large numbers right here. In any case it reduces to 6 over 25 which
is point 24. But notice again, you see the common thread
here. What we’re always doing is counting the total
number of possible outcomes as our denominator and counting how many of those outcomes fit
into the event which we’re talking about whose probability we’re computing. But in any case that takes care of part A. Let’s move on to part B. In part B we’re looking at the cases where
both Eric and Edith get jobs. Notice the denominator doesn’t change, we’re
still giving out 6 jobs among 25 people. But now if Eric and Edith get jobs, how many
other people get jobs? That leaves only 4 jobs for the other people, among how many
other people are there beside Eric and Edith, well there are 23. So we’re giving out 4 jobs among the remaining
23 people and that can be done in 8,855 ways out of this many total ways. And here the fractional arithmetic reduces
to point zero 5. So that’s only 1 chance in 20 that both Eric
and Edith get jobs. But once again, we’re again listing the total
number of outcomes in the denominator and in the numerator, counting the number of possibilities
which fit into the event that we’re talking about, which in this case is the event that
both Eric and Edith get jobs. Let’s turn to poker. Pud is playing 5 card draw, when he picks
up the hand that’s dealt to him, what’s the probability that it will be a full house?
A full house is a very, very good poker hand. We have 2 things to count here, we have to
count the total number of possible poker hands. And that’s straight forward with the combinations
formula because we know there are 52 cards in the deck and we’re playing 5 card draw,
so we’re dealt 5 cards from the 52. Combinations formula cranks that out as something
over two and a half million possible poker hands. Now I believe in one of the earlier videos
in part 4 we also counted the number of ways to get a full house. Just to review how that was done, a full house
remember was 3 cards of one type and 2 cards of another type. For example, 3 jacks and 2 sevens. And this is what we found when we did this
calculation earlier, and the reason was because you can think of it as 13 choices fo! r which
rank card the set of 3 alike comes from like the 3 jacks times 12 possibilities for which
the other rank like 7, for example, would be where the 2 cards come from. And then you also have to take into account
that it could be any 3 of the 4 cards of the one rank. Like any 3 of the 4 jacks and any 2 of the
4 cards of the other rank. Like any 2 of the 4 sevens for instance. When you put that together with the multiplication
principle and you come out with something over 3 thousand ways that you can get a full
house. But that’s a very small portion of the total
number of poker hands. And when you divide that out, you find out
that there’s only a slightly more than a one tenth of 1 percent chance of being dealt a
full house in your initial 5 cards. Now of course in most poker hands, 5 card
draw for example. You get to give away some cards and then get
some more. So your chances of ending up with a full house
are substantially better than that. But! what we’ve calculated here is the probability
of being! dealt that in your initial 5 cards. But again we’re illustrating again that because
of all poker hands being equally probable, it’s just a counting problem. We count the total number of possible poker
hands and put that on the bottom. And then we have to count the number of hands
that constitute a full house and we put that in the numerator. That’s the common thread when you’re working
with situations where all outcomes are equally likely. Mary has written some postcards. 4 are going to California, 6 are going to
South Carolina but then after she did that, 4 of the cards have been lost. And she’s worrying about the probability specifically,
what’s the probability that all the cards lost were addressed to South Carolina? Or
secondly, what’s the probability that 3 of the lost cards were addressed to California? What is it we need to count here? We need
to count the total number of possibilities for which cards were lost. We know that 4 cards have been lost out of
the 10 postcards that she wrote. She wrote 4 to California, 6 to South Carolina. She wrote 10 cards 4 of them have been lost. The number of possibilities is 210 as far
as which cards were lost. So this is our total number of possible outcomes
for lost cards, which is what will go in our denominator. Now in our first question, what’s the probability
that all the cards lost are going to South Carolina? Remember we have 6 cards going to
South Carolina, in order for all 4 lost cards to come from that group, the 4 lost cards
have to come from those 6 and the combinations formula says it’s 15 ways that that can happen. So that’s 15 out of the 210 possible outcomes
would be cases where all the lost cards are going to South Carolina. So that gives us the probability that we’re
looking for, for! the first question. For the second question where we’re looking
at the probability that exactly 3 of the lost cards were addressed to California, No change
in the denominator, since we’re still talking about 4 lost cards from among the 10 that
were written. Now we’re looking at the case where 3 of the
lost cards came from the 4 addressed to California, and one of them comes from the 6 that were
addressed to South Carolina. The combinations formula gives us the number
of ways each of those things can happen. We multiply it because of the multiplication
principle and we need to have both of those things happen in order to be in the event
we’re discussing. So we come out with 24 ways that it could
happen that 3 of the lost cards would be to California and 1 of them to South Carolina. That’s 24 out of 210 possible outcomes. A school teacher is doing safety patrol duty
or rather assigning children to do safety patrol duty. The children are Eddie, Bob, Mike, and Alice. They work on 4 different streets, Elm, Park,
Van Dyke, and Oakwood. What’s the probability mike will be assigned
duty at Elm Street? What we’re doing here is matching up the 4 students with the 4 streets
and that’s an ordering problem, it’s a matter of how many different ways can we order the
4 students so as they match the 4 streets. So the number of possibilities is going to
be 4 factorial. Think of it as just having a list of the names
of streets and then putting the name of one of the children in each spot. 4 factorial possibilities now the event we’re
talking about is Mike being assigned to Elm Street. If we match Mike up with Elm, that leaves
3 factorial ways of matching up the other 3 children with the other 3 streets. Just think about it as having 3 blanks and
we have to fill in th! e other 3 names in the other 3 blanks. So that’s 4 factorial ways altogether but
matching Mike with Elm, that leaves us only 3 factorial ways of matching the others. So we have 4 factorial possible outcomes but
only 3 factorial outcomes in the event whose probability we’re computing. So this reduces to one fourth as our probability. What’s the probability Mike gets Elm Street
and Alice gets Oakwood? Total number of possibilities is unchanged. You’re still matching up 4 children with 4
streets. 4 factorial ways of doing it but if we match
Mike up with Elm and Alice up with Oakwood, then that leaves only 2 other children and
2 other streets and there are only 2 factorial ways ofdoing that remaining matching so that
leaves us only that many ways of doing things so that these 2 conditions are met. That’s 2 of the 24 possible ways probability
now is one twelfth. Finally what is the probability Mike gets
Elm, Alice get Oakwood and Bob gets Van Dyke? Well if you’re going to match things this
way then that leaves only Eddie and Eddie would have to get the only remaining street
which is Park Street so there’s only 1 way to do it. If you’re going to meet all these conditions
and that’s 1 of the 4 factorial, 1 of the 24 possible outcomes so probability one twenty-fourth. 6 married couples are attending a party. 2 door prizes are going to be given away to
2 different people. We have 2 questions. One asks what’s the probability that a married
couple wins both prizes? So for example, Mr. Smith might win 1 prize and Mrs. Smith win the other one. Or second, what’s the probability that 1 prize
goes to a man and the other to a woman? For the first question we know that we’re
giving away 2 prizes among 12 people. So 2 of the 12 people will win prizes. Let’s count the how many possibilities there
are with regards to which 2 people win the prizes. That’s simply a matter of specifying 2 of
the 12 as the prize winners. The combinations formula gives us that number
12 times 11 divided by 2 is 66. That’s the total number of possibilities for
which 2 of the people who are at the party win the prizes. So that’s going to be our denominator. What we have to count for our numerator is
how many ways it can happen that a married couple wins both prizes. Now stop and think how many married couples
do we have? Don’t make this harder than it is. We have 6 married couples, so that means if
a couple is going to win the prizes, there are 6 possibilities for which couple wins
the 2 prizes. It could be the Smiths; it could be the Jones
and so forth. 6 married couples, 6 po! ssibilities for which
couple wins the prizes if the prizes are to be won by a married couple so that’s 6 out
of the 66 total which gives us a probability of one eleventh. Let’s pursue this a little bit further just
to get it to be intuitively appealing to you. Think about the prizes as being given out
1 at a time. We’ve got these 12 people and let’s say they
draw the prizes by drawing out of a hat or something and they’ve given out the first
prize. So we know who won the first prize. Now remember the prizes have to be won by
2 different people so how many people are eligible for the second prize? There are 6
married couples there so that’s 12 people, so that leaves 11 eligible people for the
second prize. And of those 11 people how many of them are
married to the person who won the first prize? One of them, right? So of the 11 remaining
people who are eligible for the second prize, only 1 of them is the spouse of whoever won
the first pr! ize. So that again, would say 1 chance in 11, that
the prize ! winner for prize number 2 is the spouse of the person who won prize number
1. So I’m simply suggesting another way to think
about it that should be intuitively satisfying to you and which confirms the answer that
we just calculated. Now let’s move on to part B, what is the probability
that 1 prize goes to a man and the other to a woman? Of the 66 possible outcomes we have to count
how many of those cases would be the case where 1 prize winner’s a man and the other
a woman. So for the 2 prize winners to split that way
the prize winners have to include 1 of the 6 men and 1 of the 6 women, the # of ways
to specify 1 of the 6 men as a prize winner, is 6. The number ways to choose 1 of the 6 women
as a prize winner is 6. Multiply because of the multiplication principle
and you get 36 ways to pick a man and a woman as a prize winner remember that’s 36 out of
the 66 total for a probability of 6 over 11. Would you like to hear an intuitively satisfying
way of seeing why 6 over 11 is the right answer? Let’s go back and think again about the prizes
being awarded 1 at a time. If you’ve given out the first prize, there
are 11 people eligible for the second prize. How many of them are the same sex as the person
who won the first prize? 5 right? So that means of the people eligible for the ! second
prize, 5 of them are same sex as the person who won the first prize. 6 of them are the opposite sex of the person
who won the first prize, so that’s 6 chances in 11 that prize winner number 2 will be the
opposite sex of prize winner number 1, which confirms the calculation we did by a quite
different way using the combinations formula already. Lots of times people divide up so as to be
able to play some kind of game. You want 2 teams of equal size. So let’s suppose we have 10 people dividing
up to play basketball. In this instance 2 of the 10 are brothers. And our first question we’re going to consider
is what are the chances the brothers will be on the same team, if the players divide
up arbitrarily. And then we’ll come back later and presuming
one of the people in the group is a friend of theirs named Max, what’s the probability
that all 3 of them will be on the same team? But let’s start just with the 2 brothers what’s
the chances the 2 brothers wind up on the same team? Let’s think of the teams as Team A and Team
B and let’s focus on team A and think about how many possibilitiesthere are for what players
are on team A. There are 10 people. Team A will be made up of 5 of the 10 people
so the combinations formula will give us how many different possibilities there are for
which 5 of the 10 wind up on team A. Now in the numerator, let’s count how many
possibilities there are to fill out team A in such a way that the 2 brothers are on that
team. If the 2 brothers are going to be on that
team then they will occupy 2 of the 5 spots on that team and 3 of the other 8 people not
counting the 2 brothers. 3 of the other 8 people will be the other
players on that team. So combinations of 8 choosing 3 would be the
number of ways to fill out the team starting with the 2 brothers and picking 3 of the other
8. So this fraction right here, total number
of possibilities in the denominator and number ways to fill out team A built around t! he
2 brothers to start with. This would give us the number of ways to make
up team A that includes the 2 brothers. But we have to keep in mind the question simply
says; what’s the probability the brothers will be on the same team? So we have to realize
that this would also work out with the brothers on the same team if the 2 brothers were on
team B reather than team A. That would give us another term that’s the
same so we can just take the term that we already have and multiply times 2. Or alternately, if you wanted to you could
think about the possibilities the 2 brothers on team A or making up team A so that neither
of the brothers is on it, in which case that’s the same as saying both the brothers are on
team B. In any case it’s important to realize when
we’ve done this first calculation, that’s only half the cases so we need to multiply
by 2 to have all the cases where the 2 brothers are on the same team. Now when you do this, t! otal amount of arithmetic
here you get four ninths. The alter! native calculation that’s shown
here in brackets. Denominator is number of ways of filling out
team A without any restrictions on how you do it. In the numerator, I’m thinking, filling out
team A to include the 2 brothers by including both of the 2 brothers and 3 people from the
remaining 8 people and then the reason for doubling is the same reason that there was
for doubling over here. Now let’s throw in the friend Max into the
mix. What’s the probability the brothers and their
friend Max will all be on the same team? So what’s going to be different about this calculation
from the calculation in part A? We still have this many ways of filling out team A. If we’re going to include the brothers and
Max on team A then that will make up 3 of the players and we’ll just be choosing 2 of
the other 7 players as the remaining 2 players on team A. So this fraction would represent the probability
if team A is made out randomly the probability that both brothers and Max wind up on team
A. And then we’d also have to throw in the probability
for them all winding up on team B which is why we have to double our answer for it to
be correct. Then the alternative way of thinking about
it, shown on the right here, is analogous to the alternative way of thinking about the
first calculation. We’re now going to scramble the letters in
the word mammal. Write them out in a row, what’s the probability
that the 3 M’s come first? What’s the probability that none of the M’s appear side by side?
2 questions. Think about taking the 6 letters of which
there are 3 M’s, 2 A’s and an L. And dropping those 6 letters down into these
6 spots. We learned in part 4 how to count the number
of possible ways of doing that. You could first think about which 3 slots
get occupied by the 3 M’s. So pick 3 of the 6 spots for the letter M. Then pick 2 of the remaining 3 spots for the
letter A, the 2 copies of A. and then put in the remaining spot the one
L. So doing this arithmetic with the multiplication
principle used in combination with the combinations formula, we count 60 ways to arrange those
letters from the word mammal in a row. That will give us the denominator for our
calculation. Now the question is what’s the probability
the 3 M’s come first. If the 3 M’s are to come first, then they
have to occupy these 3 spots. Which means we’ll have M, M, M in those spots. And then the 2 A’s and the L have to occupy
the other 2 spot! s. Pick which 2 of those 3 spots are occupied
by the A’s and the L will occupy the other spot. That can be done in 3 ways, so that really
only comes to 3 of the total of 60 possible outcomes. Which give us the 3 M’s in the first 3 spots. So that’s a possibility of 1 over 20. So now let’s move down to part B and see what
the solution for part B looks like. We’re considering the event that none of the
M’s appear side by side. Counting the total number of possibilities
is the same as before. We’ve already counted that there’s 60 possible
ways to arrange the letters. So now we need to count how many of those
possibilities are there for which none of the M’s would appear side by side. So for example here is one arrangement that
would give us and arrangement with none of the M’s side by side. Let’s try to list all the ways we can have
them as far as the positioning of the M’s. The M’s could occupy those 3 blanks, or these
3, or these 3 or theses 3. And I believe you’ll conclude that that’s
all the possibilities for where the M’s can go. If none of the M’s are going to appear side
by side then the M’s have to go in one of these sets of blanks. Now once we have placed the M’s, we have to
put the 2 A’s in 2 of the re! maining 3 spots and the L in the last spot. Combinations of 3 things choosing 2, this
counts the number of ways of putting in the 2 A’s once we know where the M’s go. And this is of course, just 1. The L has to go in the remaining spot. The 4 here is simply the 4 different ways
of placing the M’s that are shown. So the rationale here is, 4 possibilities
for which arrangement there for the M’s times 3 ways of positioning the A’s once the M’s
have been placed. And then only 1 remaining L. So that gives us 12 possible ways of distributing
the letters so that none of the M’s are side by side. And that’s 12 out of the 60 total for a probability
of one fifth. A shelf has 15 books. 4 of them are travel guides, 5 are biographies
and 6 of them are fiction. If you randomly pick 3 books off the shelf
what’s the probability they’ll all be fiction? We’re picking 3 of the 15 books so the denominator
will simply be the number of possibilities for that being done; picking 3 books from
the 15. The event whose probability we’re computing
is the event that all the books chosen are fiction. 6 of the books are fiction, so we’re looking
at the event where the 3 books come from the 6 fiction books. So once again, total number of possibilities
for picking 3 books from the 15 divided into total number of cases where the 3 books would
be from the 6 fiction books; which is the event whose probability we’re trying to compute. What’s the probability 2 of them will be travel
guides and 1 will be a biography? No change in the total number of possibilities,
we’re still picking 3 books from 15. Now for the numerator we need to count how
many possibilities there are if we are picking 2 of the travel guides, and 1 of the biographies. Multiplication principle comes in because
we’re doing both of those things. What’s the probability you select 1 book from
each of the 3 genres? We have 3 types, travel, biography, fiction. What’s the probability that you when you pick
3 books randomly off the shelf, what’s the probability that you get one of each type? No change in the total number of possibilities
now we’re counting the number of ways we can pick 1 travel guide, 1 biography and 1 fiction. Again tying it all together with the multiplication
principle because the event we’re looking at requires each of those to happen. What’s the probability all 3 will be the same
genre. That is 3 travel guides, 3 biographies, or
3 fictions. Think of it as 3 separate problems. The chances of getting 3 travel guides would
involve counting the number of ways we can pick 3 from the 4 travel guides. Here’s the probability of getting 3 travel
guides. Similarly, here’s the probability of getting
3 biographies. And finally this is the probability of getting
3 of the fiction books because we would be picking 3 from the 6 fiction books. So just do those computations separately and
then combine them, add them together. 4 ways for getting 3 travel guides plus 10
ways for getting 3 biographies, plus 20 ways for getting 3 fictions, gives us a total of
34 possible ways of getting 3 books from the same genre. And so our probability is 34 over 455. And finally what’s the probability that 2
of the books will be from one of the categories and the remaining book will be from a different
category? Let’s think of all the ways that could happen. We could get 2 travel books and a biography,
2 biographies and 1 travel, 2 travels and a fiction, 2 fictions and 1 travel, 2 biographies
and 1 fiction, or 2 fictions and 1 biography. That’s it. If we’re going to get 2 from 1 category and
1 from the other category then 1 of these things has to happen. Now in fact this one, we calculated as 1 of
the earlier steps in this example. So one way of doing the problem would be simply
to calculate the probability of these other 5 cases in a similar manner and then add up
the 6 probabilities to get the answer. Little bit long winded, but nothing particularly
difficult about doing that. There is another, and in fact, substantially
quicker approach however that I would like to remind you of, and that is to simply keep
in mind that this is bound to happen unless you have all 3 books from the same category
or else you have books from the 3 different categories. In other words, no matter what 3 books you
get, their either going to be all from the same category or 2 from 1 category and 1 from
another category, or else 1 book from each of the 3 categories. This is one of those 3 cases and we have already
computed the probabilities of the other 2 cases. So we can get the probability of this case
by simply doing 1 minus the probabili! ties of the other 2 cases. 1 minus the probability of all 3 being from
the same category which we’ve already calculated. And also subtract off the probability of them
being from 3 different categories. So this is in effect making use of the fact
that the probability of the complement of an event is 1 minus the probability of an
event. The complement of the event we’re looking
at here is the union of these 2 other events that we’ve already talked about, whose probability
we’ve already calculated. These 2 previous events are mutually exclusive
so the probability of their union is just the sum of their probabilities so subtracting
off the probability of the union of the 2 previous events is just a matter of subtracting
off each of them 1 at a time.

## One Reply to “Uniform Probability Distributions (movie 5.3)”

1. Bobby Walker says:

Boring……..