in this lesson we want to look at some probability word problems so at some point in your algebra course or your pre-calculus course you're going to come across a small section on probability and i just want to go through some basic word problems that you're going to see and talk about how we can solve these guys so i'm just going to jump right in i'll just kind of explain things as we go it's a very very easy topic overall so jason flips a coin and then rolls a six-sided die what is the probability that the coin lands heads up and okay i'm going to highlight that and the die shows an even number okay so when you deal with these problems you're going to have problems that say and so some event occurs and some other event occurs in this case one event is flipping a coin and i'm getting a heads up and the other event is rolling a die and getting an even number so a 2 a 4 or 6. okay when you hear the word and you need to be thinking about multiplication so i would take and use this formula the probability of some event a and some event b happening okay in this case we'd say that coin landing heads up and also the die showing an even number is equal to the probability of this event a happening times the probability of this event b happening okay now this is given the fact that they are independent events meaning the first event does not impact the result of the second event in any way shape or form okay if he flips a coin that does not impact the result of rolling a six-sided die okay so these are independent events we can just use this formula as it's given so what we want to do here is just think about the fact that the probability for flipping a coin and getting a heads up is going to be what it's going to be a half or 50 percent we all know this because we've thought about that throughout life at some point flipping a coin saying hey my odds are 50 50. well how do we show this mathematically where does that 50 50 or the 50 percent come from well essentially what you do is you think about the number of different ways that you can get an outcome from flipping a coin well you can basically get a heads up or tails up right so there's two possibilities so i'll put that in the denominator and then think about what you're looking for what's the result well i want a heads up and that can only happen one way right so i would put that number in the numerator okay and as we do more of these this will make more sense but essentially this is the formula you want to use that's where one half comes from this is basically the number of favorable outcomes in this case a favorable outcome would be landing heads up over the total number of possible outcomes and the possible outcomes here would be landing heads up or landing tails up okay so i know that's one half so let's just say that this satisfies the probability of event a now what about the probability of event b let's say that's the die showing an even number well with a die if it's 6 sided it can land on a 1. let me just kind of write this down here in a different color a 1 a 2 a 3 a 4 a 5 or 6. so we know that the 2 is an even number the 4 and the 6. okay so 2 4 and 6 those are even numbers so how many different favorable outcomes can i have well i have 1 two three favorable outcomes that are possible out of a total of six so i would put three over six which you can simplify to one half okay and then you just multiply the two together and let me just kind of erase this you multiply the two together and you get 1 4. okay now you can leave it as a fraction and say it's one-fourth for the probability or you can convert it to a decimal which is 0.25 and then to a percentage and say it's 25 percent okay so essentially the probability of this json guy flipping a coin and getting a heads up and also taking a die and rolling it and getting an even number is 25 percent okay let's look at a slightly more challenging problem and in this case we're going to have dependent events so allison a teacher out of middle school has five boys and six girls in her class allison randomly selects three different students to walk up and give a presentation once the presentation is over the student will leave for the day what is the probability that the first student is a boy the second student is a girl and the third student is also a girl so you see the problem here is that these events are not independent okay once she picks a student and we're looking for her to pick a boy first okay then the next time we want her to pick a girl and then the next time we want her to pick a girl but each time she's picking the class size is changing right so the number of boys is changing the number of girls is changing whatever it is in that scenario but the class size is constantly changing so we have to readjust our probability okay so we're still going to think about multiplying the probabilities together but you use a different formula for this if it was just two things here it's three but let's say it was two things so the probability of a and b okay if these guys are dependent it's just two it would be the probability of a times the probability of b given that a has happened okay so given that a has happened what's the probability of b happening okay and in this case we're going to adjust it because we have one more thing to consider because now we're thinking about the probability that the first student is a boy okay so that happens then the probability that the next student is a girl okay and then the probability that the third student is also a girl so again you have and so you're thinking about multiplication okay so what is the probability on the first graph that that student is a boy well there's going to be five boys okay so there's five possible choices in there that's going to be boys out of a total of 5 plus 6 is 11 11 total students so the first number we're looking at is i'm just going to write this down here 5 over 11. i'm just going to put this as the first okay so the second is going to be what well if i go back up i know that the student leaves he can't be re-chosen now and there's going to now be what well there's one less boy so there's four boys and there's the same number of girls but the total number in the class now is four plus six or ten okay so if i think about picking a girl now it's going to be six girls out of a 10 student classroom okay let me just erase this so let's go down here and put that now it's six over ten okay so this is second you'd say and now what's going to happen for the third choice when we go back up well now there's let me just kind of scratch this out again say this is four this is going to go down to five there's a total of nine students and there's five girls so there's five over nine okay let me just erase this again and now i'll say this is five over nine and this is going to be the third pick okay so the probability of picking a boy on the first run is five over eleven then given the fact that that event has already happened now there's a six over ten probability of picking a girl now given the fact that both of those events happened the probability on the third pick that it's a girl is five over nine okay so again to get the final probability or the probability that all these events happen so this and she picks the first student as a boy and she picks the second student as a girl and she picks the third student as a girl it's going to be the product of these probabilities so before we multiply notice you can cancel this with this and get a 2 here and then 6 divided by 2 would be 3 and then 9 divided by 3 would be 3 okay so i'm going to put a 1 here and basically what you have is 5 times 1 or 5 over 11 times 3 which is 33 so you have 5 over 33 which is the probability of picking a boy then a girl then a girl all right so now let's talk about some word problems that involve this or so when event a happens or an event b happens so with this we're going to be adding now so when you hear and you've got to be thinking about multiplication when you hear or you've got to be thinking about adding okay but there's a trick to this and let me just kind of go through this with this problem so katie has 6 nickels and seven dimes in her pocket five of the nickels and one of the dimes are canadian the others are from the us suppose katie randomly selects a coin from her pocket what is the probability that it is a dime or is from the u.s this is the key word here the or okay so what you want to be thinking about is the probability of some event a happening or some event b happening is the probability of the event a happening plus the probability of the event b happening minus okay this is important minus the probability of a and b happening okay i know a lot of people use set notation there so you might see this as well okay so the intersection of a and b however you want to write it doesn't really matter i'm just going to stick to writing and i think that's a little bit easier for us to understand okay in this particular case we don't have something known as mutually exclusive events mutually exclusive events are two events that can never occur simultaneously so in this particular case we don't have that because it says what is the probability that it is a dime or is from the u.s well we have dimes that are also from the u.s so we have to use this full formula okay if we didn't have dimes from the u.s we could knock this part out okay this is going to take out the double counting so let's go through real quick and think about what's the probability that it's a dime well we know that she has six nickels and seven dimes okay well six plus seven is 13. so 13 coins overall seven of them are dimes so the probability of having a dime is 7 over 13. then plus what's the probability okay that the coin is from the u.s meaning it's not canadian well it tells you that five of the nickels and one of the dimes are canadian so six out of the 13 coins are canadian so 7 out of the 13 are american or u.s currency so 7 over 13. 7 plus 7 is 14 14 over 13 is a number larger than one when we deal with probabilities it's either zero okay meaning it's not going to happen up two and including one one means it's certain to happen anything in between means it may or may not happen okay but you can't get a probability less than zero or greater than one that doesn't make any sense okay so that would be a red flag if you forgot to subtract this out so all we need to do to fix this is subtract out the probability of a and b occurring okay so how many dimes do we have that are also from the u.s well let's think about that we know that there are seven dimes in her pocket okay total of seven and it says that five of the nickels and one of the dimes are canadians so that means that six of the dimes are from the u.s okay so that tells me that i need to subtract away 6 over 13 right because 6 of the dimes are from the u.s so if i crank this out now i get 7 plus 7 which is 14 and then 14 minus 6 which is 8 so i get a result of 8 over 13. so that's going to be the probability that she picks a coin that's going to be a dime or it's from the u.s okay let's look at another example of this and now we're going to see mutually exclusive events so beth has a fruit basket which contains four apples four peaches and five pairs if she randomly selects a piece of fruit what is the probability that it is an apple or a peach again when you see this keyword or you're thinking about addition okay but in this particular case you don't need that full formula okay because it's not possible for her to select something that's an apple and a peach at the same time so again when you think about the probability of some event a happening or some event b happening it's the probability of a plus the probability of b and then minus the probability of a and b occurring and i didn't write my a there so a and b well in this particular case you can get rid of this because it's going to be zero again you cannot have a piece of fruit that's an apple and a peach at the same time maybe they're working on that i'm not sure but in this scenario it's not possible so we just think about the probability of a let's say that's where she selects an apple well there's four apples out of a total of four plus four plus five four plus four is eight equals five is thirteen so the total of four over thirteen okay that's going to be your probability of picking an apple then plus what's the probability of picking a peach well again it's 4 over 13. so all i have to do is just sum these and i get 8 over 13. all right so let's wrap up the lesson by looking at an example where we use our binomial probability formula so the desks in a classroom are organized into four rows and four columns each day the teacher randomly assigns you to a desk you may be assigned to the same desk more than once over the course of five days what is the probability that you are assigned to a desk in the front row exactly three times okay so if you are dealing with a scenario such as this where you have repeated independent trials okay so this is independent because every day you're randomly assigned to a desk and you can sit where you've sat before and the outcome in each trial is either a success meaning in this case you would get assigned to the front row or a failure okay so if you want to find the probability of our successes in n trials you can use this formula so let me write this down here and we'll come back up in a minute so n choose r okay we remember that and then times p raised to the power of r p is the probability of a success so on a single day what's the probability of you getting seated in the front row i'll get to that in a second then times 1 minus this p raised to the power of n minus r so n is the number of trials so if we go back up we know that this is going to happen for five days so over the course of five days so n is five and the number of successes we're looking for is r right so three times as we're looking for that number to happen so n is five n is five and r is three okay so you're just plugging things in so you have five and three okay then p is the probability of a success on any given day well if we go back we know that there's a total of 16 deaths right four rows and four columns four times four is 16. so in the front row you're going to have four desks so that means you have a four over 16 or one fourth or 25 percent probability of getting seated in the front row on any given day so we're going to put a 1 4 here you could use a decimal if you want it doesn't matter it's going to be the same result in the end r again is 3 and then 1 minus 1 4 if you're using a fraction you would do 4 4 minus 1 4 which is 3 4 and then raised to the power of n which is 5 minus r which is 3 so this is going to be squared okay so once you have this set up you can just erase this and crank this out so on a calculator 5 choose 3 is going to be 10. then if we think about 1 4 cubed that's going to be 1 over 4 cubes of 64. then if we think about 3 4 squared 3 squared is 9 4 squared is 16. can i cancel anything before i multiply i just get rid of this i don't need anymore can i cancel anything well yeah i can cancel a factor of 2 here and here so this would be 32 this would be 5. so 5 times 9 is 45 over what is 32 times 16 that's going to be 512. so 45 or 512 that's going to be your probability of getting seated in the front row exactly three days out of a total of five days