hi i'm erica you may recognize me from some of my other youtube videos on the gmat
now these days i mostly spend my time working on sat and a ct here with magoosh but today i've
got a great gmat tip on probability permutations and combinations that'll help all of you who are
aiming for a 700 plus gmat score so probability permutations and combinations are gonna show up
at about the same frequency no matter what level you're performing at on the test what changes
with your difficulty level is the complexity of those questions now oftentimes with these
more complex problems there are so many moving pieces so many things to separately solve for
that solving for all of them and then combining them together to get the total of what you want
is unrealistic in the scope of two minutes so what do you do so on a problem where you're finding
it's particularly complex to solve directly for what you want instead consider solving for
what you don't want and subtracting it from the total now again this tip applies to probability
permutations and combinations but let's start by discussing how it applies to probability not only
does this appear more frequently in probability questions there's also a couple of things that
make it a little bit easier in probability questions so the first thing that we should
know about probability questions is that the total probability is always one or one hundred percent
so we don't need to solve for that separately that is something we already know and then the
second part here is that there is often a clue in the question and that's the phrase at least
if we see the words at least in a probability question that is a huge hint that we should
be using this trick so let's take a look at some example questions to explore this so this question
asks us to find the probability that at least one multiple of three is rolled now let's talk about
what at least one means well at least one could mean exactly one so we could roll a multiple of
three on our first roll or we could roll it on our second roll or we could roll it on our third roll
or on our fourth roll and if we were to add up the probabilities of each of those cases
we would get the total probability that we roll exactly one multiple of three but that's not what
we're looking for we're looking for at least one which means that we could also roll exactly two
multiples of three which again can happen in any order first and second first and third first and
fourth and so on but then we can also roll exactly three multiples of three again in any order
or we could roll all four multiples of three now if we wanted to find each of those cases
find the probability for each one and add them up that would work for this problem but it takes a
long time and it's really error-prone let's instead try our trick let's take our total
probability of one and subtract the probability of what we don't want well what don't we want we want
at least one multiple of three so the only thing we don't want is rolling no multiples of three
now that's a significantly easier probability to find so on our first roll we can roll a one
a two a four or a five so that's four options of six or a two-thirds probability of not rolling a
three now on our second roll we have the exact same options a one a two a four or a five out
of six options so that's four out of six or again two-thirds same thing for our third roll and same
thing for our fourth roll now since we need all four of these we multiply them together
to get 16 over 81. now be extremely careful because that's an answer choice note that if
you are using the strategy of solving for what you don't want and subtracting from the total
they will very frequently put what you don't want as an answer choice
so we have to remember that to turn it into what we do want we need to subtract from that total or
subtract from one now in this case 1 minus 16 over 81 gives us 65 over 81 or answer choice d
let's try one more example probability question before we move on to permutations and combinations
all right so this problem gives us a bag of marbles and wants us to find the probability that
a blue marble is taken from the bag at least once so once again we have that clue with at least now
if we didn't catch that we would see in the answer choices that we also have a one minus in most of
the answer choices so we've got an abundance of clues that we want to use our strategy of taking
the total probability one and subtracting the probability of what we don't want now if we didn't
notice this we might get hung up on trying to find the probability of one blue marble or two blue
marbles or three blue marbles or four blue marbles and then deal with all the different orders that
come from it but because we've seen these clues we're going to jump straight to our strategy
of our total probability one minus the probability of what we don't want well what don't we want well
if we want at least one blue marble the only thing that doesn't work is no blue marbles
now once again this is a much easier probability to find so we have eight picks
out of the bag all right now for the first pick we want not a blue don't care if it's a red
don't care if it's a green that's irrelevant to this problem but we have 10 non-blue
marbles out of 15 total marbles or a 2 out of 3 probability all right now notice here that the
marble is replaced between each draw so we have with replacement which means that for this second
pick we are still going to have 15 marbles and still 10 of them are going to be non blue so we're
gonna have the same probability on the second pick of two thirds and we're gonna have that same
two-thirds probability for each of the eight picks now this could be pretty complicated math but
we're going to go ahead and again take the q from our answer choices which has a bunch of exponents
right so because this has a bunch of exponents we're going to leave this in exponential form
since there are eight picks and each pick is two-thirds probability we have two-thirds to the
eight now we're lucky here that's not an answer choice few because we have to remember to subtract
from one to go from what we don't want to what we do want so one minus two thirds to the eighth
is answer choice c so this is obviously a powerful tool for probability and if you stopped the video
right here you would have learned something very valuable but what makes this strategy
so cool is that it also has amazing applications for permutations and combinations questions where
there are difficult restrictions let's take a look at a permutations question to illustrate so this
problem presents us with a permutation where we have a gift that is wrapped in some kind of paper
that is tied with some kind of ribbon and affixed with some kind of bow now all of
these are different blue paper is different than a blue ribbon is different than a blue bow
so this is again a permutation now we know that we have five different colors of wrapping paper
red green blue purple or gold four different colors of ribbon red blue gold or silver and
four different colors of bow blue red purple or gold now if the question wrapped up here by asking
how many different possible selections li jing could make for wrapping a particular present
we would use the fundamental counting principle to multiply these up and get that we have 80 total
options for wrapping this present but there is an extra sentence in here where it says the paper
ribbon and bow cannot all be the same color so there is a restriction now at this point we could
try to find all of the different permutations where all of them are different or where the paper
and the ribbon are the same or where the paper and the bow are the same or where the ribbon and the
bow are the same and we could add all of those up and that would work but what we might notice
here is that there is only one thing that's not allowed which is that they're all the same color
so why don't we use our rule again we'll take our total number of options for wrapping our package
or our present and then subtract the number of options where they're all the same color
all right now we already have our total there are 80 different ways so let's go ahead and figure out
how many ways we can wrap this present where all of the wrapping items are the same well we
know that we could have red wrapping paper a red ribbon and a red bow we can have blue wrapping
paper a blue ribbon and a blue bow or we can have gold wrapping paper gold ribbon and a gold
bow and that's it there are only three options for them all being the same all red all blue we're all
gold so we can take our total of 80 subtract our three options where they're all the same color
to get our correct answer of 77 where they are not all the same all right this last question is
a combinations question and it is very difficult probably more difficult than anything you'd see
on the actual test but it really illustrates the power of this strategy so this question asks us to
find a combination of 3 out of 15 dots that do not form an equilateral triangle so kind of a giveaway
that we have a combination here and we have a pretty obvious restriction not an equilateral
triangle so we could waste a bunch of time here trying to draw a bunch of non-equilateral
triangles but we should notice our answer choices are huge so that's not going to get us very far we
can try to think about all the different types of triangles that aren't equilateral but again that's
going to be kind of complicated when we should notice that there is a very specific restriction
it's not equilateral so what we can do instead is use our trick find our total combinations
of three that make a triangle and subtract the number of triangles that are equilateral
all right so let's start with our total we have 15 dots we're choosing 3 so that's a 15 choose 3
combination so that's going to be 15 factorial over 3 factorial times 15 minus 3 factorial or
12 factorial all right pulling 12 factorial out of 15 factorial we get 15 times 14 times 13
over three times two times one all right the one doesn't really mean anything pull the three out
of the fifteen to get five the two out of the fourteen to get seven so we've got five
times seven times thirteen there are several different ways to approach that but that's
not what this video is about so i'm going to tell you that this is 4 55. all right now at
this point this is our total we still need to subtract the number of equilateral triangles
so the combinations that give us an equilateral triangle so we know that our answer is going to be
smaller than 455 which means c d and e don't make any sense now if we were to guess blindly from
here we would have a 50 chance of getting this question right but before we do that and before
we start drawing triangles let's think about this logically do we think that there are going to be
more equilateral triangles so triangles where all three sides are exactly the same length
or more triangles where the sides can be any length as long as they're not all the same
seems likely that it's going to be the second right the equilateral triangle is a very specific
type of triangle so there's likely to be fewer of them now let's take a look at our answer
choices in light of that if our answer was a 160 that would mean there were about 300 equilateral
triangles to 160 non-equilateral triangles that doesn't fit our expectation at all on the other
hand if we were to pick b 450 that would mean there were only five equilateral triangles
compared to significantly more 450 non-equilateral triangles that fits our expectation significantly
better so we can go ahead and pick b and roll out a now if you want to prove this to yourself you
can draw the triangles and if you want to see that done that is in a blog article that i've
gone ahead and linked in the description but i would not do that on test day here we can get to
the answer by using our trick and a little bit of logic and that's a 700 plus gmat probability
permutations and combination strategy instead of solving for what you want consider solving for
what you don't want and subtracting from the total to stay updated with all of our upcoming videos
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