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Course: IIT JEE > Unit 1
Lesson 1: IIT JEE- Trig challenge problem: arithmetic progression
- IIT JEE perpendicular planes (part 1)
- IIT JEE perpendicular plane (part 2)
- IIT JEE complex root probability (part 1)
- IIT JEE complex root probability (part 2)
- IIT JEE position vectors
- IIT JEE integral limit
- IIT JEE algebraic manipulation
- IIT JEE function maxima
- IIT JEE diameter slope
- IIT JEE hairy trig and algebra (part 1)
- IIT JEE hairy trig and algebra (part 2)
- IIT JEE hairy trig and algebra (part 3)
- Challenging complex numbers problem (1 of 3)
- Challenging complex numbers problem (2 of 3)
- Challenging complex numbers problem (3 of 3)
- IIT JEE differentiability and boundedness
- IIT JEE integral with binomial expansion
- IIT JEE symmetric and skew-symmetric matrices
- IIT JEE trace and determinant
- IIT JEE divisible determinants
- Intersection of circle & hyperbola
- Common tangent of circle & hyperbola (1 of 5)
- Common tangent of circle & hyperbola (2 of 5)
- Common tangent of circle & hyperbola (3 of 5)
- Common tangent of circle & hyperbola (4 of 5)
- Common tangent of circle & hyperbola (5 of 5)
- Trig challenge problem: multiple constraints
- Trig challenge problem: maximum value
- Vector triple product expansion (very optional)
- IIT JEE lagrange's formula
- Representing a line tangent to a hyperbola
- 2010 IIT JEE Paper 1 Problem 50: Hyperbola eccentricity
- Normal vector from plane equation
- Point distance to plane
- Distance between planes
- Challenging complex numbers problem: complex determinant
- Series sum example
- Trig challenge problem: system of equations
- Simple differential equation example
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IIT JEE complex root probability (part 2)
2010 Paper 1 problem 31 Math (part 2).avi. Created by Sal Khan.
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- how are there only six ways. the question says r1 r2 and r3 are 3 different turns.
so r1 could be= 1,2,3,4,5,6
r2= 1,2,3,4,5,6
r3=1,2,3,4,5,6
a same number could occur thrice so he dint take that part in consideration.
there should be a lot more possibilities.(3 votes)- First i too thought bout that but.. you should consider that there can be no repititions for eg..if r1=1 then r2 cant be 1 again..etc etc.. so effectively to fill each r you can use only 2 values so that the sum comes to 0..but if in other case r2 becomes 1 then r1 must be something else...if we go on we get 6 cases like this by varying 1 and 4 with r1,r2 and r3..hence at the end he found out the probability of 1 case and multiplied it with 6..hope you understand my geeky language..if not i can still explain by solving the whole question with steps..(1 vote)
- Why should w^2 = conjugate of w? Is it a special case or a theorem?(0 votes)
- w and w^2 are both roots of equation x^2 + x + 1=0. And if the roots of a quadratic equation are complex they are always conjugates of each other.(6 votes)
- if 15 boys of different ages are distributed into 3 groups of 4,5 and 6 boys randomly , then probability that three youngest boys are in different groups is what!(1 vote)
- 3*2*1*4*5*6*12!/15! = 6/91. the explanation is:
The youngest person can be filled placed in one of the groups in 3c1 ways
Second person can be placed in any of the remaining two groups in 2c1 ways.
The third person can be placed in 1c1 ways.
The remaining boys can be arranged in 12p12 = 12! ways.
Now, in the first group among 4 positions, the younger (chosen person from above) can be in 4 places.
Similarly, in second group the chosen member can be filled in 5 ways.
In third group, the chosen member can be filled in 6 ways.
So, the total favourable permutations are 3c1*2c1*1c1*12p12*4*5*6
the total permutations are 15p15 = 15! [ Here I am talking arrangements or permutations of events, but it would not be wrong in this case IN THIS CASE, to take even combinations of events]
The probability is 3*2*1*12!*4*5*6/15! = 6/91(3 votes)
- What does unity mean?(1 vote)
- the maximum possible number of points of intersection of 7 straight
lines and 5 circles is what!(1 vote)- 1 circle can intersect one line in 2 ways .5 circles in 10 ways.for 7 lines 70 ways.
now 7 lines intersect each other in 21 ways by nC2.
total points is70+21 that is 91(1 vote)
- why the probability for r1, r2, r3 can't be found out by 6C2(combination)?
what's the wrong thought on it? plz help!(1 vote)- Why are you choosing 2 objects/number sides from 6 objects/number sides ?
During the event of r₁, there are 6 possible outcomes [1, 2, 3, 4, 5 or 6] with the event happening once. You will see exactly one face of the die (i.e. one occurring of the rolling event). You won't roll it six times!(1 vote)
- In the last part shouldn't we divide (1/27) by 6, since the probability of the required arrangement is 1/6?(1 vote)
- In the last part shouldn't we divide (1/27) by 6, since the probability of the required arrangement is 1/6? (7:45)(1 vote)
- In the last part shouldn't we divide (1/27) by 6, since the probability of the required arrangement is 1/6?(1 vote)
- Can't you just do r1 r2 and r3 have to be 1,2,3 mod 3? Then there are 2 choices (1,4) (2,5) (3,6) for each and 2 cubed is 8. Now you can arrange them in 6 ways to get 8 times 6 which is 48 desired outcomes over 216(6 cubed) total outcomes. This comes out to be 2/9.(1 vote)
Video transcript
In the last video, we started
on this problem right here, where we said let omega be a
complex cube root of unity, and omega cannot be equal to 1. And so we figured out what
all of the complex cube roots of unity were. We figured out-- well, we
knew that 1 was one of them, and we used that to factor
out this third degree equation right over here. And we figured out the other
roots, cube roots, of unity were negative 1 plus or minus
the square root of 3i over 2. And we said, look, the problem
said let omega be one of the, or be a complex
cube root of unity. So I just picked it to
be the negative version. So I said let omega be
this thing right over here. Now, just to kind of explore
the space a little bit, we said, OK, what's
omega squared? And we figured out that it was
actually the conjugate of this. It's the plus version of this. So we got omega squared
right over here. And then I kind of
wasted your time a little bit because we
know what omega cubed is. We know that omega is
a cube root of unity, so omega cubed must be unity. But it wasn't harmful, I guess,
to go through the process. It shows you sometimes
my brain gets into a rut and just does stuff that
it doesn't have to do. But we figured out-- we
multiplied this times omega to show that, hey, it is
definitely equal to 1. So what we were able to
do is set up a situation, and we will use this to
think about the next part of the problem. So omega is equal
to negative 1/2. Or maybe I should say
omega to the first power is equal to negative 1/2 minus
the square root of 3 over 2i. Omega to the second
power-- let me write it over here--
omega to the second power is equal to negative 1 plus
the square root of 3 over 2i. And then omega to
the third power, and that was pretty
straightforward, is equal to 1. And I touched on this
in the last video. What is omega to
the fourth power? Omega to the fourth
power is going to be omega to the
third times omega. So it's going to be omega again. So this is also
going to be-- let me scroll to the
left a little bit-- this is also going to be
omega to the fourth power. Omega to the fourth
power is the same thing as a omega to the first
power, which is this. Now, what's to the fifth power? Well, it's going to be omega
times the fourth power, so it's omega times this. Well, omega times this is
the same thing as omega squared, so it's
going to be that. So this is omega
to the fifth power. And what's omega
to the sixth power? Well, that's just the same
thing as taking omega cubed and squaring it, so this
is also going to be 1. So this is omega to the sixth. Or you could just use,
it's this times omega because it's to the
fifth power, and we saw that that's also equal to 1. And the reason why I went
up to the sixth power is because, if you'll
remember the old problem, we're rolling a die. And on that die--
I'm assuming it's a normal six-sided
die-- we're going to get values between 1 and 6. And we're going to take omega
to the different powers, and we're going to
see if it equals to 0. And we want to find
the probability of this being equal to 0. So let me write this down. We want to find the probability
that omega-- and we're just using the omega that we
picked because they said it's one of the complex
roots that isn't 1-- so we want to find the
probability that omega to the first die of the roll
plus omega to the second die of the roll plus omega
to the third die of the roll, that when you take their
sum, that that is equal to 0. This is what we
need to figure out. So to figure this
out, let's just figure out what combinations
of powers of omega will even add up to 0. How can we even
add them up to 0? I mean, they all have
to cancel out some way. And if you look
at them, it looks pretty interesting of how
they might cancel out. If I take one version
of this, the only way that they really
can cancel out is if I take one version of
this, and add it to this, and then add it to this. Let me show you. If I take negative 1/2 minus
the square root of 3 over 2i, and I add it to this, and I
add it to this, negative 1/2 plus the square root of 3 over
2i, and then I add it to 1, and then I add it to this,
and then I add it to this, what do I get? Well, you're going
to have this guy and this guy are
going to cancel out. Negative 1/2 plus
negative 1/2 are going to be equal to negative 1. You add negative 1 to 1,
it's going to be equal to 0. So essentially, what
it was asking us, what's the probability that I
get, for each of these terms, that I get one each of each
of these powers of omega? Or another way of thinking
about this, what's the probability-- so if
we think of it this way. Let's think of it this way. r1 could be equal to-- we're
going to get omega here, which is the same thing
as omega the fourth power. So r1 could be 1 or 4. And this is in the situation
where the first one gives us omega. The second one is omega
squared, or the second one gives us this value. And then the third
one gives us 1. So r2 would be 5 or 2. And then r3 would
be equal to 3 or 6. Now, I want to be very clear. We could swap these around. There's actually six ways
that you could-- instead of this being the first roll,
this roll could be 1 or 4, and then this one
could be 5 or 2, and then this one
could be 3 or 6. So there's actually
six ways of doing this, I guess you could say. You could permute these six
different ways, or 3 times 2 times 1. But we'll think about
that in a second. But assuming that we want this
way, where this first one is going to evaluate to
this, and the second one is going to evaluate to
this, and the third one is going to evaluate
to this, what's the probability
of that happening? And then we're going to multiply
that 6 by 6 because there are six ways to arrange
these terms right over here. So let's do that. So what's the probability
of that happening? Well, the probability
that r1 is a 1 or a 4, well, that's two
values out of six, so the probability is 1/3 there. The probability
that r2 is a 5 or 2, well, that's also
going to be 1/3. There's two values out
of a possible of six. The probability that
r3 is a 3 or a 6, only two possibilities there. So two out of six possible
faces of the die, so times 1/3. So the probability
that this first one is going to evaluate to this
value right over here, the second one is going
to evaluate to this value, and the third one is
going to evaluate to 1, is going to be 1/3 times 1/3
times 1/3, which is 1/27. And I touched on this
already, there's six ways. You could rearrange
this six ways. You have three terms, and you're
putting them in three places. So in the first place, you
could put three of the terms. In the second place,
you have two terms left that you could put. And in the last place, you
only have one term left. So there's 3 times 2 times 1
ways to arrange these things. We can arrange this 3 times
2 times 1 different ways, so there's six
different arrangements. The probability of each is 1/27. So the probability under
question, this thing over here, there's six arrangements. There's six arrangements of
getting these things to be added up in this way, and the
probability of each of them is 1/27. So it's equal to
6/27, and if you divide the numerator and
the denominator by 3, it's equal to 2/9. And we're done. It wasn't too bad, I think. The probability of getting
omega to the r1 plus omega to the r2 plus omega to
the r3, where r1 and r2 and r3 are numbers obtained
from rolling a fair die, the probability--
when you add all of these, it's going to be
equal to 0-- is 2/9. And I thought that was
a pretty neat problem.