Examples of Binomial Distribution, Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com

hi welcome to this tutorial on the
binomial distribution this is the third
one in the series and in this I'm going
to show you how to calculate various
probabilities from binomial
distributions now I'm assuming that
you're familiar what a binomial
distribution is and also how to work out
the probability of our successes if not
please go back and look at my first two
tutorials now for the first question
we've got a manufacturer of a bag of
sweets claims that there is a 90% chance
that the bag contains some Toffees and
if 20 bags are chosen what is the
probability that one all the bags
contain Toffees and impart two more than
eighteen bags contain Toffees now this
particular question is a binomial
distribution why because there are a
finite number of trials there's 20 in
this particular problem
each referring to a particular bag of
sweets there are two outcomes in any
trial success failure success being
where we pick a bag that has Tuffy's in
it failure being that the bag does not
have Tuffy's the probability of success
remains constant all the way through at
naught point nine and the probabilities
are independent of one another in any
trial so typical binomial distribution
but I'm not encouraging you to draw a
tree diagram I just want you to be aware
that this is what's going on in the
background okay well let's just remove
the tree diagram or we'll get on with
the question now the first thing we need
to do is to define a random variable so
it can be any letter I'm going to choose
X and make sure you write a capital
letter for a random variable I'm going
to let X be the random variable and I'll
write RV for short
and it's going to represent the number
of bags with Toffees okay so we just
write that in don't be lazy always write
the description of what your random
variable is going to say if it doesn't
say it in the question and we have to
define the distribution that it comes
from so in this case where X is
binomially distributed remember it's two
parameters for the binomial distribution
the number of trials which is in this
case 20 for the sweets and next is the
probability of success that is a bag
contained Tuffy's and the probability
that happening is not 0.9 for 90% and
the first question probability that all
the bags contain Tov is so for part one
we're looking for the probability that x
equals 20 and assuming that you're
familiar then with the formula it will
be 20 see and then the number of
successes which in this case is 20 then
we write down the probability of success
which is not 0.9 and this occurs with a
success rate of 20 so that's to the
power 20 and then you've got naught
point 1 the probability of failure and
that would occur not know x next we need
to work this out on the calculator and
you should find that you get naught
point 1 2 1 5 7 and so on which when
rounded comes out at naught point 1 to 2
to 3 decimal places so that's a very
simple basic question working out the
probability of a particular value and
the next example though we've got
several values because we're being asked
the probability that the number of bags
that contain Toffees is more than 18 so
X is more than 18 now the random
variable X
takes on the values zero because you
could find you've got no bags containing
Tuffy's or one bag could contain Toffees
two bags and so on all the way up to 20
bags so if we're looking for X to be
more than 18 it could be that we have 19
bags that contain Tuffy's or it could be
20 bags that contain Toffees and we'd
have to add these two probabilities
together and if we now use the formula
for the probability x equals 19 it will
be from 20 trials choose 19 successes
probability of success is not 0.9 and
that's repeated 19 times over
probability of failure is not 0.1 and
that would only occur once then we've
got plus the probability of x equals 20
and we could either use the former again
or we could just simply use the answer
we had previously not 0.12 1 5 7 and so
on and if we add these two values
together what you should find is that
you get naught point three nine one
seven four and so on which is the same
as naught point three nine two to three
decimal places 3dp okay let's have a
look at another question now in this
question in clinical trials a certain
drug has an 8 percent success rate of
curing a known disease and if 15 people
are known to have the disease what is
the probability that at least two are
cured lissa again is a typical tree
diagram we've got a finite number of
trials 15 this 2 outcomes in any trial
that is that the drug is successful or
it fails the probability of success is 8
percent and it remains constant
throughout the problem and in any trial
the probabilities of success are
independent of the previous trial
so a typical binomial distribution so
first of all to do this let us define a
random variable can be any letter
remember so now why don't we choose Y
okay let Y be the random variable in
this particular problem and it will be
the number of people cured so we define
that and we now need to say where Y is
distributed binomially two parameters
number of trials first which is fifteen
and the probability of success next
which is eight percent or not point zero
eight so in this particular question
we're asked that the probability that at
least two are cured so that's the
probability that Y is greater than or
equal to two now in this particular
problem y can be no people being cured
or one person being cured 2 people 3 4
etc all the way up to and including 15
so this is the same as the probability
that y equals 2 plus the probability
that y equals 3 y equals 4 and so on all
the way up to the probability that y
equals 15 now this would be a horrendous
um to work out and there is a way around
this problem what we should know is that
if we were to work out the probability
that y equals not plus the poverty y
equals 1 plus the probability y equals 2
3 and so on all the way up to 15 that
that would total 1 all the probabilities
would total one so since I'm not using y
equals 0 and y equals 1 I could take
those probabilities away from 1 and it
would give me this hour
sir so I could do one minus the
probability that y equals nought and I'd
need to put this in a bracket plus the
probability that y equals 1 these are
the only other situations that I've not
included here so all I need to do now is
just work out these two rather than
having to do all this slot so b1 minus
so for the probability y equals not it
would be 15 C not probability of success
is not 0.08 and that is occurring no
times probability of failure is not 0.92
and if we had no successes there must
have been fifteen failures now we work
out poverty y equals one so that B plus
fifteen C one probability of success not
0.08 would have occurred just once and
the probability of failure would have
occurred fourteen times so you have
naught point nine two to power 14 and if
I leave it to you to work this sum out
remember I must put that in brackets by
the way if I leave it to you to work
that some out you should find that you
get one - not point six five nine seven
and so on and if you take that away from
one you end up with naught point three
four zero to seven and so on which when
rounded to say three decimal places is
not point three four zero to three
decimal places now that brings us to the
end of this tutorial and I hope you've
been able to follow these examples and
you can apply them in similar questions
now in my next tutorial what I want to
show you though is how on some occasions
we can do questions like this by using
commutative binomial
tribution tables and that can save us an
awful lot of work so I hope you'll have
a look at that tutorial
you