# 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)”

Boring……..