Class 12 Exam > Class 12 Videos > L24 : Random variables : Examples-2 - Probability, Maths, Class 12
L24 : Random variables : Examples-2 - Probability, Maths, Class 12 Video Lecture
FAQs on L24 : Random variables : Examples-2 - Probability, Maths, Class 12 Video Lecture
| 1. What is a random variable in probability? |  |
| 2. How are random variables useful in probability? | |
Ans. Random variables are useful in probability as they allow us to mathematically model and analyze uncertain events. By assigning probabilities to different outcomes, we can calculate the likelihood of certain events occurring and make informed decisions based on those probabilities.
| 3. Can you provide an example of a random variable? | |
Ans. Yes, for example, consider rolling a fair six-sided die. The random variable X can represent the number that appears on the top face after the roll. X can take on values 1, 2, 3, 4, 5, or 6, each with a probability of 1/6.
| 4. What is the difference between a discrete and a continuous random variable? | |
Ans. A discrete random variable can only take on a countable number of distinct values, such as the number of heads obtained in multiple coin flips. A continuous random variable, on the other hand, can take on any value within a certain range, such as the height of a randomly chosen person.
| 5. How can random variables be used to calculate probabilities? | |
Ans. Random variables allow us to define probability distributions, which specify the probabilities associated with different values of the variable. By summing or integrating these probabilities over certain ranges, we can calculate the probabilities of specific events or conditions involving the random variable.
Text Transcript from Video
hello friends this video on travel tip
our 24 is brought you by exam feel calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 to power 23 find the
provided education of a number of heads
in to toss of a coin this Kaunas tossed
twice so you may get head or tail in
first and then he receiving it Headley
that day so logically you have head head
head tail tail head these four options
we have right
so you toss two coins you get these many
options so we have to find the value
addition of number of heads so number of
heads can be 0 can be 1 and can correct
so X can have values 0 1 2 why because I
am looking for here the random variable
X is nothing but the number of heads
please note the random variable X has
value 0 1 into because I'm looking for
the number of heads here
right so the Verity of X is equal to 0
is number of head is 0 is TD this
probability of getting tail tail that is
1 by 4 correct because there are 4
possible options one of these is still T
so why aren't you getting X equal to 1
is nothing but either you get h t or t h
HT t or th this is nothing but two
options
the parity of getting X equal to do that
is the number I have heads twice is
nothing but sovereignty of getting
traded
there is nothing but 1 by both say we
see the distribution the value of X can
be 0 1 & 2 and the stability of X can be
higher
1 by 4 4 in case of 1 it was 1 by 2 case
also it was and if you add all these if
you add this becomes 1 where who
correct
so nothing big deal here so you have
this two vanities right so you are
looking for a random variable number of
heads we saw that number of heads can be
0 1 or 2 so we formed x is 0 X 21 X / -
OH - take my exam and find the Coverity
distribution of the number of tails in
this case number of tails in toss of
three coins so you talk three coins you
see the toss three coins you can get
head or tail in the first then you may
get head T then you may get red team in
second and third also here we get ahead
theme you may get a head team here also
you may get a head tail NOC here we get
a head then eight possible options one
two three four five six seven eight let
me jot down like head head head head
head tail head tail head head tail tail
tail head head tail head tail tail tail
head tail tail tail eight possible
options I have so I am looking for
number of teens so number of T is zero
here number of days is one here number
of theta is one here / update is 2 here
1 - number of TL - number of these we
see the value of x can be the number of
tails can be either 0 1 2 or 3 right 1 2
or 3 so prepare D of X equal to 0 so
there are only one possible options 1 by
8 so Verity of X is equal to 1 that is
the number of tail is 1 is 1 2 3 3 white
well if your accident - is one two three
I mean three by and the very name of
number of teens that is X is going to
the for three years one by eight only
one possible option so we see now the
distribution will be like this in the
value of x can be zero one and dreams
and your TX can be 4 0 it was 1 by P for
1 and s 3 5 K for 2 to 3 by 8 and 4
wonderful
and if you add also you see that 1 by 8
plus 3 by 8 plus 2 that a blasphemer a
deliverer and that's why it's a random
find the probability distribution of
number of heads in for toss of a coin
very similar machine very similar the
last question
they it was three coin heads for coin so
you solve in same approach I want to
solve this volume you can example you
have this first point can be head team
second coin can also be head tail head
tail third kind of camp coin and so
every head tail head tail head tail head
tail now I can see the fourth coin can
also mean head tail head tail
this guy can also be head tail head tail
right
this guy can also be head tail this guy
can also be head tail head teen so these
are 16 possible scenarios 1 2 3 4 5 6 7
8 and 18 years right so 16 possible
scenarios then you count number of heads
scenarios where you get head you'll find
that head is equal to 1 in one scenario
sorry 0 in once in are you it is equal
to 1 you will find in let's suppose four
scenarios head is equal to who you'll
find is six narrows here is equal to
three you'll find in four scenarios head
is equal to 4 again you'll find out
once that is so the quality of this guy
will be 1 by 16 Israeli four by sixteen
6 by 16 by 16 that's all you'll get you
try to solve not that difficult you have
since you're looking for the number of
heads so X is nothing but number of
heads and X can take value 0 1 2 3 now
actually find parity of X equal to 0 1 2
3 4 you find these values let's take one
more example so here you define the
provider distribution of the number of
success in to throw up a die this guy's
throne tries the success is defined
number greater than 0 oh so you throw
this die you may get both the numbers
you turns right twice right t1 t2 in
both the case you may get number lesson
4 right in one case you may get number
less in whole or in both the case you
may get number it so player 3 option
right so now X is nothing but number
greater than 4 so X in half arrows 0 1
into correct path because my in both
trial you may get number less than 4 so
there's 0 possible success in one case
you may get number greater than 4 one
possible in both case you may get number
greater than 4
correct so in any throw so variety of
number greater than 4 so we're ending up
in greater than 4 this is nothing but
guarantee of getting 5 6 and that is
nothing but you why since and the Verity
of number less than equal to 4 is
nothing but a variety of reading 1 2 3
pull your vessels correct so for X is
equal to 0 this parity is nothing but
priority of n less than 4 I mean I'll
write like this actually let's suppose
this guy is success and this guy is
failure
like number given for success I feel
here there's nothing but rivality of
failure failure do get Fillion bookies
so this becomes 4 by 6 into 4 by 6 that
is my reading of X is equal to 1 means
you either get failure in first access
in second or you get success in first
failure in second for vanity of and
there are independent events so I'm just
writing me like this
so the vanity of failure into success is
nothing but vanity or failure is what
will basics
well it's a business 2x6 last now why do
your success is to boy 60 when do plated
is four vessels that is nothing but for
money
similarly X is going to do there's
nothing but you get success both case
that means you are looking for 2 by 6
into 2 by 6 that is so you can see the
pilot distribution you can value of X
can be 0 1 2 and the probability values
for 0 is 4 by 9 or what also for my 9:00
in for two also remaining and that is my
test so if you add all the times 4 + 4 +
8 + 1 9 9 - 1 so that's how it is for
number 3 - difference X this is the case
now second part was 6 appears at least
in one die
so let me give a space for this more
space for this question okay I'll try to
squeeze in get some speed yeah so now
when you are getting 6 at least in one
dies next right so X and I'm going to
value say 0 1 because either it may have
6 in one time or it may not happen
correct
c6 getting 6 at least in one die is my X
X is nothing but getting 6 in one
cluster right one or more than one so
either it may not get or it may get so
let's remove this is why extreme ache
don't make it X you can choose so I can
have values
0 & 1 correct
so where do your Y is equal to 0 is
there is no 6
there is no 6 you got right so come and
pick off getting 6 is probability of
getting 6 is on my 6 and priority of not
getting success by basis correct but my
6th and not getting this is one minus
one person that's five basic so when you
are saying that you don't get any sex
that means you are saying that coburn
you're getting 6 - six - that is nothing
but Phi by 65 versus since they are
independent events correct there is
nothing but 25 by 36
so your liberal idea of getting once
this thing that is hilarity of getting
at least one side so you get you don't
get 6 you get 6 plus you get six in the
first throw or you may get 6 in motion
correct because it is at least here at
least one die so at least one tie
satisfy at least one this one guy one
guy all the time
so this is nothing burns then they are
dependent 20 of 6 not as 5 by 6 forgot
name is 6 is 1 by 6 last vanity of 6 is
1 by 6 into 6 notice 5 by 6 plus 20 of 6
is 1 by 6 in the I independent event
right I can write this as 1 2006 in 2006
2006 in 2006 not over T of 6 not anybody
who says this you solve is there
something but this guy you see the value
of X can be 0 and 1 only and which me
uc40 came out to be 25 by 36 and for
what it means and if we are both you get
26 by that is acceptance
oh okay thank you
visit exam viacom to watch free
educational videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more
thanks and serene
our 24 is brought you by exam feel calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 to power 23 find the
provided education of a number of heads
in to toss of a coin this Kaunas tossed
twice so you may get head or tail in
first and then he receiving it Headley
that day so logically you have head head
head tail tail head these four options
we have right
so you toss two coins you get these many
options so we have to find the value
addition of number of heads so number of
heads can be 0 can be 1 and can correct
so X can have values 0 1 2 why because I
am looking for here the random variable
X is nothing but the number of heads
please note the random variable X has
value 0 1 into because I'm looking for
the number of heads here
right so the Verity of X is equal to 0
is number of head is 0 is TD this
probability of getting tail tail that is
1 by 4 correct because there are 4
possible options one of these is still T
so why aren't you getting X equal to 1
is nothing but either you get h t or t h
HT t or th this is nothing but two
options
the parity of getting X equal to do that
is the number I have heads twice is
nothing but sovereignty of getting
traded
there is nothing but 1 by both say we
see the distribution the value of X can
be 0 1 & 2 and the stability of X can be
higher
1 by 4 4 in case of 1 it was 1 by 2 case
also it was and if you add all these if
you add this becomes 1 where who
correct
so nothing big deal here so you have
this two vanities right so you are
looking for a random variable number of
heads we saw that number of heads can be
0 1 or 2 so we formed x is 0 X 21 X / -
OH - take my exam and find the Coverity
distribution of the number of tails in
this case number of tails in toss of
three coins so you talk three coins you
see the toss three coins you can get
head or tail in the first then you may
get head T then you may get red team in
second and third also here we get ahead
theme you may get a head team here also
you may get a head tail NOC here we get
a head then eight possible options one
two three four five six seven eight let
me jot down like head head head head
head tail head tail head head tail tail
tail head head tail head tail tail tail
head tail tail tail eight possible
options I have so I am looking for
number of teens so number of T is zero
here number of days is one here number
of theta is one here / update is 2 here
1 - number of TL - number of these we
see the value of x can be the number of
tails can be either 0 1 2 or 3 right 1 2
or 3 so prepare D of X equal to 0 so
there are only one possible options 1 by
8 so Verity of X is equal to 1 that is
the number of tail is 1 is 1 2 3 3 white
well if your accident - is one two three
I mean three by and the very name of
number of teens that is X is going to
the for three years one by eight only
one possible option so we see now the
distribution will be like this in the
value of x can be zero one and dreams
and your TX can be 4 0 it was 1 by P for
1 and s 3 5 K for 2 to 3 by 8 and 4
wonderful
and if you add also you see that 1 by 8
plus 3 by 8 plus 2 that a blasphemer a
deliverer and that's why it's a random
find the probability distribution of
number of heads in for toss of a coin
very similar machine very similar the
last question
they it was three coin heads for coin so
you solve in same approach I want to
solve this volume you can example you
have this first point can be head team
second coin can also be head tail head
tail third kind of camp coin and so
every head tail head tail head tail head
tail now I can see the fourth coin can
also mean head tail head tail
this guy can also be head tail head tail
right
this guy can also be head tail this guy
can also be head tail head teen so these
are 16 possible scenarios 1 2 3 4 5 6 7
8 and 18 years right so 16 possible
scenarios then you count number of heads
scenarios where you get head you'll find
that head is equal to 1 in one scenario
sorry 0 in once in are you it is equal
to 1 you will find in let's suppose four
scenarios head is equal to who you'll
find is six narrows here is equal to
three you'll find in four scenarios head
is equal to 4 again you'll find out
once that is so the quality of this guy
will be 1 by 16 Israeli four by sixteen
6 by 16 by 16 that's all you'll get you
try to solve not that difficult you have
since you're looking for the number of
heads so X is nothing but number of
heads and X can take value 0 1 2 3 now
actually find parity of X equal to 0 1 2
3 4 you find these values let's take one
more example so here you define the
provider distribution of the number of
success in to throw up a die this guy's
throne tries the success is defined
number greater than 0 oh so you throw
this die you may get both the numbers
you turns right twice right t1 t2 in
both the case you may get number lesson
4 right in one case you may get number
less in whole or in both the case you
may get number it so player 3 option
right so now X is nothing but number
greater than 4 so X in half arrows 0 1
into correct path because my in both
trial you may get number less than 4 so
there's 0 possible success in one case
you may get number greater than 4 one
possible in both case you may get number
greater than 4
correct so in any throw so variety of
number greater than 4 so we're ending up
in greater than 4 this is nothing but
guarantee of getting 5 6 and that is
nothing but you why since and the Verity
of number less than equal to 4 is
nothing but a variety of reading 1 2 3
pull your vessels correct so for X is
equal to 0 this parity is nothing but
priority of n less than 4 I mean I'll
write like this actually let's suppose
this guy is success and this guy is
failure
like number given for success I feel
here there's nothing but rivality of
failure failure do get Fillion bookies
so this becomes 4 by 6 into 4 by 6 that
is my reading of X is equal to 1 means
you either get failure in first access
in second or you get success in first
failure in second for vanity of and
there are independent events so I'm just
writing me like this
so the vanity of failure into success is
nothing but vanity or failure is what
will basics
well it's a business 2x6 last now why do
your success is to boy 60 when do plated
is four vessels that is nothing but for
money
similarly X is going to do there's
nothing but you get success both case
that means you are looking for 2 by 6
into 2 by 6 that is so you can see the
pilot distribution you can value of X
can be 0 1 2 and the probability values
for 0 is 4 by 9 or what also for my 9:00
in for two also remaining and that is my
test so if you add all the times 4 + 4 +
8 + 1 9 9 - 1 so that's how it is for
number 3 - difference X this is the case
now second part was 6 appears at least
in one die
so let me give a space for this more
space for this question okay I'll try to
squeeze in get some speed yeah so now
when you are getting 6 at least in one
dies next right so X and I'm going to
value say 0 1 because either it may have
6 in one time or it may not happen
correct
c6 getting 6 at least in one die is my X
X is nothing but getting 6 in one
cluster right one or more than one so
either it may not get or it may get so
let's remove this is why extreme ache
don't make it X you can choose so I can
have values
0 & 1 correct
so where do your Y is equal to 0 is
there is no 6
there is no 6 you got right so come and
pick off getting 6 is probability of
getting 6 is on my 6 and priority of not
getting success by basis correct but my
6th and not getting this is one minus
one person that's five basic so when you
are saying that you don't get any sex
that means you are saying that coburn
you're getting 6 - six - that is nothing
but Phi by 65 versus since they are
independent events correct there is
nothing but 25 by 36
so your liberal idea of getting once
this thing that is hilarity of getting
at least one side so you get you don't
get 6 you get 6 plus you get six in the
first throw or you may get 6 in motion
correct because it is at least here at
least one die so at least one tie
satisfy at least one this one guy one
guy all the time
so this is nothing burns then they are
dependent 20 of 6 not as 5 by 6 forgot
name is 6 is 1 by 6 last vanity of 6 is
1 by 6 into 6 notice 5 by 6 plus 20 of 6
is 1 by 6 in the I independent event
right I can write this as 1 2006 in 2006
2006 in 2006 not over T of 6 not anybody
who says this you solve is there
something but this guy you see the value
of X can be 0 and 1 only and which me
uc40 came out to be 25 by 36 and for
what it means and if we are both you get
26 by that is acceptance
oh okay thank you
visit exam viacom to watch free
educational videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more
thanks and serene
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