Examples : Binomial Distribution Video Lecture | Mathematics (Maths) Class 12 - JEE

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things now so here a fair coin is coin
is taunt toss 10 times I know that throw
up coins of our knowledge trial so here
n is equal to take 10 times find the
probability of getting exactly six hits
I am looking for the probability of six
heads right so let head is equal to
success so I am looking for robbery of
six success by formula this is nothing
but n see X where X is equal to 6 I'm
looking for the parity of X is equal to
6 success and see X 3 to the power X Q
to the power n minus X let's put the
value n is 10 X the 6 so value of
getting head is 1 by 2 like successive 1
by 2 into T bar 6 failure is also 1 by 2
here n is what 10 minus 6 if we solve as
you get 1 0 5 by 5 to the sync part at
least at least 6 hits at least success
that means I'm looking for 6 7 8 9 10 it
right there's probability of 6 head
that's what I do
7 head 2008 heads when you 9 heads plus
10 this is what I'm looking for so 6
heads if you see is nothing but 10 n is
10 C 6 C this guy will always come out
to be 1 1 by 2 to the power 10 because P
is 1 by 2 Q by 2 if you saw this becomes
NC X 1 by 2 to the power X 1 by 2 to the
power n minus X right this becomes 1 by
2 to the power n we cool things we got
NC X into 1 by 2 to the power 10 right
plus 7 head
see 7 1 by 2 to the power 10 + 10 c8 1
by 2 to the power 10 + 10 c9 1 by 2 to
the power 10 + 10 10 why I took this
because you can do this way also 1 by 2
to the power 6 into the middle before
but everywhere it came out to be over
but important
this is only in this example because
head and tables are symptomatic so if
you solve this you will get 193 by that
makes this at most six hits that means I
am looking for probability of 1 2 3 4 5
6 heads
same thing here looking for poverty of
one hit or 0 head alright 0 it also
parity of zero hit osmolarity of 1 head
where do you have to hit vanity of 3
head over into your forehead so where do
you have 5 it where do your 6 here right
you add this 0 headers 10 C 0 1 by 2 to
the power 10 you keep adding this
poverty or 6 head will be 10 C 6 1 by 10
power 10 you add all these values you
get 50 3 by 64 advanced super only you
just apply the binomial distribution to
find all these things no you toss coins
and number of times ten times hundred
nine thirty times and then you find the
probability that you got six heads or
you probably you got at the most excels
those kinds of things easy to find using
the gamma distribution
take for my example ten eggs are drawn
successfully with replacement false not
contain ten percent defective X so this
there is a defective there is ten
percent you have fresh eggs find the
probability that there is at least one
defectively
so let since we are looking for
detective egg so let's suppose success
is defective so no matter your so this
will be 10 percent
that is one Bible and temerity of
failure will be 1 minus 1 by 10 does my
little correct we are looking for
defective only so I assume that surface
is defective
there is at least one defective a and
they are 10 X drawn so n is equal to 10
so preventing of at least one success is
nothing but 2001 success to success 3
success you keep adding till 10 or you
can do smart way you can see that one -
where it a of 0 sources correct who the
same you can do this way also in that
case you have to add so idea one success
to success 3 success for success or
smart you can say that this is nothing
but 1 minus dou by dou Y of 0
subsistence correct so a value of 0
success what n is 10 10 C 0 1 by 10 is
the probability of success to the power
0 9 by 10 is if earlier failure to the
bar 10 and what you get is 1 minus 9 to
the power 10 by 10 and the disaster very
simple application of binomial
distribution plus a smart way of handing
this position where I could have done
this way I could have find one success
to success in some sense I can say that
this is nothing but 1 minus 2 pi 0 0 6
so PN of die is thrown four times so n
is equal to 1 on this one and it giving
a doublet is success so success is
getting a doublet that is 1 1 2 3 3 4 5
this is the success so provided your
success will be what 6 by 36 is 1 by 6
plus 6 into 6 36 combinations right we
have to find the probability of to
success so first let's find the value of
e 0 so that will be 1 minus 1 by 6 that
is 5 assets so I am looking for a
variety of
do success that will be if you see
probability of let's suppose and success
or a success is nothing but or X success
is over small NC x peeled apart x QT bar
and minus X similarly for X is going to
do you get n is 1024 is 4 for x for X is
2 for C 2 P was successes 1 by 6 1 by 6
to the ball XLS to failure is 5 by 6 to
the power n minus 2 is 4 minus 2 that is
4 minus 2 so if you solve this you get
25 by X and the disaster very simple
question all right so just use the
Bernoulli's trial to get answer the say
for example they have 5 percent
defective item in large bulk of their
some items out of this defective is
equal to 5 percent what is the
probability that sample of 10 items that
means you have 10 bundl is trap will not
include more than one defective item so
let's say success is equal to defective
item so success is getting detective ID
so in this case property of success will
be 5 percent 500 that is 1 by 20 and the
word your failure will be 1 minus 1 by
20 that is 19 by train correct so I am
looking for a case where not more than
one item added there is I am looking for
either 0 item are defective or one item
added at the max these two correct so
I know the formula that polarity of X is
small X is nothing but and see X weakly
power X 3 bar in magazines so 4 X is
equal to 0 it will be N and this is what
1010 see is 0 well do success is 1 by 20
the power 0 failure is 19 by 20 Libba 20
minus 0 so 10 minus 2 plus 4 X equal to
110 C 1 T is 1 by 20 to the power 1 and
19 by it will you Cooper lady - 1 so if
you solve this you get 29 by 20 into 19
by 2050 power of mine and that is my
answer
let's take one example five cards are
drawn successively with just the
placement of a well stuffed 52 cards
what is the probability that all 5 cards
are spades if you see here it is with
replacement so it's all independent
event so we can use the Bernoulli's
trial here
all 5 cards are speed so we can say that
light successes spell card right so no
guarantee of success is nothing but
getting spade is what 13 by completing
this one better
and the reality of failure will be 1
minus 1 by 4 that is 3 so I've got too
many do success and failure I am looking
for all 5 card or speed because I'm
looking for X 5 this will be n CX that
is fine there will be a five card are
drawn 5 C 5 successes of my four Billy
got 5 period is 3 by 4 to the power Phi
minus Phi and this value is nothing but
1 by 1 0 position correct
second is only 3 cards are spent same
thing Rivera T of X equal to 3 here this
becomes n 5 C 3 1 by 4 to the power 3
and 3 by 4 to the power 5 minus 3 okay
you solve this comes out to be 45 by 5
now listen
thus you have to understand that this is
a binder wheel or you can apply the
minam in clients and then you can just
apply the form like advanced let's take
an example so here it say that the bulb
that the pilot is the bulk original
factory will fuse after one who today in
this point zero five we defined it of
Aldi that all the five such purse that
is invisible to fine none none more than
one not more than one more than one and
at least one meters after 50 days so my
success is that the Vulgate through
straight bulk use one fuse so my clarity
of success is find zero five and my
preventive failure will be one minus
this that is point Nyjah okay and I know
that property of X is equal to X is
nothing but NC X right e there is a
success to the power takes failure to
the power in minus X the first is none
against Spain that is provided a of X is
equal to 0 that is something but v c0 p
is 0.005 to the power 0 qu is 0.95 to
the power n minus 0 is 5 minus so you
solve this you get 0.95 those bodies are
not more than 1 that means either 0 or 1
that means I'm looking for X is equal to
0 plus P of X right this one so P of X
plus you know I know is 0.95 to the
power 5 P of X equal to 1 is nothing but
Phi C 1 Phi n 0 Phi to the power one
point nine five you got 5 minus 1 and if
you solve this you solve this all thing
you get point nine five to the power 4
in the third one more than one more than
one means P of X is equal to 1 2 3 4 5
that is nothing but 1 minus P of X is
equal to 0
whatever sorry more than one is start
from correct and this is nothing but 1
minus this we have already got 4 0
anyone that is fine I try to the bottom
Oh
similarly at least one at least 1 is
nothing but P of at least 1 is nothing
but P of 1 minus P of less than right do
you have less than or 1 minus P of 0
actually so P of 0 if you know that is
nothing but point 9 10 to the power 5 X
1 minus P of X producer so that's how it
is just the same question just you
understand that when you're talking
about none you read for X bill to 0 not
more than 1 that is you're looking for 0
or 1 looking for more than 1/3 looking
for 2 3 4 5 that you can write there is
nothing but 1 minus P of X below 0 and 1
and you were looking for at least one
that you are looking for 1 2 3 4 5 right
that you can buy - nothing but 1 minus P
of instability similar to this what we
have done here so it was 2 3 4 5 here it
is 1 2 3 4 5
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