# Binomial Distribution and the Classical Definition of Probability

1. Nov 7, 2014

### Soumalya

I am facing problems while comparing the results of solving a problem individually using both the concept of Binomial Distribution of Probabilities and the Classical Definition of Probability.

Let me formulate the problem first:
"The probability that a pen manufactured by a company will be defective is 1/10.If 12 such pens are manufactured find the probability that exactly 2 pens are defective."

Solution: If 'X' be a random variable representing the number of defective pens ,

then, 'X' is a bivariate with parameters n=12 and p=1/10 where q=1-p=9/10.

For exactly 2 defective pens out of 12 manufactured pens we should have the probability of such an event as,

P(X=2)=12C2.p2.q12-2=0.2301

Now,if we try to solve the problem in a different way using the classical definition of probability,

we have n(S)=2n=212=4096
and, n(E)=nCr=12C2=66, where 'E' is the event of obtaining 'r' successes in 'n' independent trials i.e, 'r' defective pens among 'n' manufactured pens in case of this particular problem.
So P(E)=n(E)/n(S)=66/4096=0.016.

Thus, from the classical definition of probability we obtain the probability of 'r' successes in 'n' independent trials of an experiment as something different from what we obtained using the theory of binomial distribution of probability.

Can anybody explain where did I go wrong?

2. Nov 7, 2014

### PeroK

In the second case (classical probability) you forgot about the probabilities of each event! You simply counted the number of ways an event can happen. So, your answer is independent of p. Which has got to be wrong.

For example, what would happen if you set p = 0 (or, at least, set p to be a very small number). Your second answer would not change.

3. Nov 7, 2014

### Soumalya

By the classical definition of probability,

P(E)=n(E)/n(S)

Now if 'E' can happen in many ways,

P(E)=number of all possible ways 'E' can happen/number of all possible outcomes of the experiment.

Isn't that correct?

If yes then if we describe 'E' as the event of a success and we wish to have 'r' successes in 'n' independent trials we can have 'r' success in nCr ways and the total number of all possible outcomes for 'n' trials is 2n.

Thus P(E)=number of ways we might have 'r' successes in 'n' independent trials/total number of possible outcomes for 'n' trials of the experiment
=number of elements in the set of 'r' successes in 'n' trials/number of elements in the sample space of 'n' trials

But yes you are correct when you say for various values of 'p' we would literally have the same probabilities of 'r' successes in 'n' trials of an experiment.

Again if you have p=q=0.5 we have the same results applying both the Binomial Distribution formula and the Classical Definition of Probability!

Suppose you toss a fair coin 10 times.Let 'X' denote the number of heads.Then p=0.5 and q=0.5 and n=10.

For r=2 by the formula of Binomial distribution,we have

P(X=2)=10C2.p2.q2=45/1024.

Again by the classical definition of probability,we have

n(E)=10C2 where 'E' denotes the event of having exactly 2 heads out of 10 tosses.

n(S)210=1024.

Thus P(E)=n(E)/n(S)=45/1024.

So the results match!

How do you explain this?
The Classical definition of probability holds only for equally probable events?

4. Nov 7, 2014

### PeroK

No. That's wrong. You can only count when the events are equally likely (like the toss of a coin). Which you've worked out above.

Let's go back to your pen, which has a probability of p = 1/10 of being defective. Suppose you only have 1 pen. What is the probability of its being defective?

n(E) = 1, n(S) = 2, so P(E) = 1/2

So, by your calculations, regardless of p, the probability a pen is defective is 1/2.

p = 1/2 is a special case, where every possibility is equally likely. For two coins: HH,HT, TH, TT are all euqally likely, so you can simply count.

For your pens: DD, DN, ND, NN (D = defective, N = not defective) are not equally likely events. So, you have to include the probability of each in your calculations, as well as counting them. This is precisely what the binomial formual does.

5. Nov 7, 2014

### Soumalya

Thank You Perok!

I am glad you helped me with my confusions I would ponder upon everything once again and come back if I have recurrent doubts!
Thanks Once again!!!!!!
Great Help!!!!!

6. Nov 7, 2014

### Soumalya

According to what you have said if in a random experiment we have 'equal probability' of all the possible outcomes of the experiment i.e, if all sample points of the whole sample space are 'equally likely' then we are allowed to calculate probability of an event (which is a subset of the sample space) by the fundamental principles of counting.

For example while choosing a card out of a pack of 52 cards the probability of the outcome of each individual card is 1/52 i.e, every card is equally likely to appear.So if we draw a pair of cards out of the deck of cards and wish to calculate the probability of the two cards being both hearts we may use the principles of counting which gives,

P(Both cards drawn are hearts)=No. of ways of selecting 2 cards out of 13 hearts/Total No. of ways of selecting 2 cards out of 52 cards
=13C2/52C2=1/17.

But in case of calculating the probability of obtaining 'r' successes in 'n' independent trials of an experiment,

n(S)=(No. of ways we might have 1 success and n-1 failures)+(No. of ways we might have 2 success and n-2 failures)+........................+(No. of ways we might have n success and 0 failures)

Thus S={set of all possible combinations of 1 success and n-1 failures,set of all possible combinations of 2 successes and n-2 failures,.................,set of all possible combinations of n success and 0 failures}

Now, we may have 1 success and n-1 failures in the form,

SFFFFFFFFFFFFFFF.......upto 'n' terms or FSFFFFFFFFFF..............upto 'n' terms or FFSFFFFFFFFFFFF................upto 'n' terms and so on.

Thus in every possible case of 1 success and n-1 failures we have the probability of the result of the combined experiment as=P(S).P(F).P(F).P(F)........up to 'n' terms=p1qn-1 where 'p' and 'q' are the probabilities of a success and a failure respectively in a single trial of the experiment.

Thus probability of each event of the set representing 1 success and n-1 failures=p1qn-1.
Similarly,probability of each event of the set representing 2 success and n-2 failures=p2qn-2
.................................................................................................................................................................................
.................................................................................................................................................................................
.................................................................................................................................................................................
probability of each event of the set representing n success and 0 failures=pnq0/SUP]

Thus every sample point of the sample space for the combined experiment has different 'likeliness' or chances of occurrence.
So we are not eligible to use the principles of counting in such cases.

Am I correct in my understanding so far?

Last edited: Nov 7, 2014
7. Nov 7, 2014

### PeroK

Yes, I think you understand this. You do always have to be careful that things really are equally likely. And, in general, with probablity problems you need to think it out carefully to make sure you are using the correct technique or formula.

P(2 hearts) = P(1st card is a heart)*P(2nd card is a heart, given first is a heart) = (1/4)(12/51) = 1/17

In the second case, you have effectively worked out the binomial formula for yourself. (You seemed to miss out the possibility of 0 successes and n failures, but otherwise that was a good analysis of the problem.)

8. Nov 8, 2014

### Soumalya

Yes I missed the possibility of 0 successes and 'n' failures.Thanks for the clarification!!!!!

Could you suggest me some good reading or textbooks on Probability for a better understanding?

I am an engineer graduate so I don't need extensive study on the subject but I would like get clear on the fundamentals very well.

9. Nov 9, 2014

### Soumalya

I went through some problems recently and here is one I wish to share that's been bothering me about the validity of the modern theory of probability:

"A family has 6 children.Find the probability of 3 boys and 3 girls."

Solution: According to the solution in my textbook,

The probability of a particular child being a boy or a girl is 1/2. (They are assuming equal probabilities for a child being a boy or a girl)

Now, if the occurrence of a boy is thought of being a success and 'X' be a random variable denoting the number of success then 'X' is a bivariate with parameters n=6 and p=1/2.

Hence,P(X=3)=6C3.(1/2)3.(1/2)6-3=6C3(1/2)6=5/16.

Now taking into consideration that they have initially assumed that the probability of a randomly picked child being either a boy or a girl is 1/2 or 3/6 simply denotes the probability of half of the children being boys(3 boys out of 6 children) and remaining half as girls(3 girls out of 6 children) should be 1.What I mean to say is that if they have assumed half of the children as boys and the other half as girls already then it doesn't make any sense to have the probability of same as 5/16 as calculated from the binomial theory of probability distribution.We already have 3 boys and 3 girls in the family!

But if we look at the problem in a different way say if we had formulated the problem in this way:

"In a school we have 50 children half of them being boys and the remaining girls.For a randomly chosen group of 6 children find the probability that 3 are boys and 3 are girls."

Solution: Here we could rightly assume the possibility of a child chosen randomly being a boy or a girl is 1/2 as half of the population are boys and the remaining half girls.

Now for a randomly chosen group of 6 children we could have several possibilities like:

1. All boys i.e, 6C6.
2. All girls i.e, 6C6.
3. One boy and 5 girls and the numbers of ways they can be arranged i.e, 6C1.
4. Two boys and 4 girls and the number of ways they can be arranged i.e, 6C2.
5. Three boys and 3 girls and the numbers of ways they can be arranged i.e, 6C3.
6. Four boys and 2 girls and the numbers of ways they can be arranged i.e, 6C4.
7. Five boys and 1 girl and the numbers of ways they can be arranged i.e, 6C5.

The number of all possible outcomes is=6C6+6C6+6C5+6C4+6C3+6C2+6C1=26+1.

Thus n(S)=26+1.

The rest of the problem could be solved as in the previous case.

If the occurrence of a boy is thought of being a success and 'X' be a random variable denoting the number of success then 'X' is a bivariate with parameters n=6 and p=1/2.

Hence,P(X=3)=6C3.(1/2)3.(1/2)6-3=6C3(1/2)6=5/16.

The precise difference between both the problems is that in the first case we had 6 children in all and we assumed half of them being boys and the rest as girls initially so according to pour assumption the possibility of 3 boys and 3 girls out of 6 children should be 1 whereas in the second problem we choose 6 children out of 50 children among whom 25 are boys and 25 are girls.So the possibility of each randomly picked children being of a particular gender is 1/2 and we are interested in finding out the probability of the occurrence of the set of 3 boys and 3 girls and all possible combinations of them.

I guess the solution in my textbook for the first problem is inappropriate at least according to the language of the problem as formulated?

By the way, am I correct when I say the number of all possible outcomes in 'n' trials of an experiment is 2n+1 when we consider only two possible outcomes in a single trial???

In my textbooks it's again 2n

I am confused as to how it's possible!

Lets say in 6 trials of an experiment if 'S' denote success and 'F' failures then the possible outcomes are:

SSSSSS i.e, 6C6=1.
SFFFFF and its 6C1 ways.
SSFFFF and it's 6C2 ways.
SSSFFF and its 6C3 ways.
SSSSFF and it's 6C4 ways.
SSSSSF and it's 6C5 ways.
FSSSSS and its 6C1 ways.
FFSSSS and it's 6C2 ways.
FFFSSS and its 6C3 ways.
FFFFSS and it's 6C4 ways.
FFFFFS and it's 6C5 ways.

FFFFFF i.e, 6C6=1.

The outcomes in bold are recurrent ones.

So we have the number of all possible outcomes =6C6+6C1+6C2+6C3+6C4+6C5+6C6=26+1.

Last edited: Nov 9, 2014
10. Nov 12, 2014

### Stephen Tashi

Soumalya,

. If we say "Half the children in a family are boys. What is the probability that a randomly selected child from that family is a boy?" then we get the answer 1/2. But if we say "The probability of randomly selected child being a boy is 1/2. What is the probability that a family of 6 has 3 boys?" this specifies a different problem because you aren't given the fact that half the children in the particular family are boys.

It's unlikely that an author would wish the statement "The probability of a child being a boy is 1/2 and a family has 6 children" to mean the definite fact that 3 of the children in the family are boys. What such a statement means could be debated, but it's a question of interpreting language, not an ambiguity in the mathematical theory of probability.

To pose a mathematical problem correctly, information about actual results such as "half the children are boys" must be stated as a definite facts. You can't inform a reader than outcome actually happened or actually happened 8 out of 10 times by only stating the probability of the event. Probabilities have no guaranteed relation to actual outcomes. This is a disturbing feature of trying to study any type of uncertainty. When we study a topic, we want to have certainty in the statements we make about it. We want theorems that say "if ...such-and-such ... then ...so-and-so is definitely true". It's human nature to seek theorems that say "if the probability of ...is such-and-such then so-and-so must definitely happen." or "If the probability of .. is such-and-such then so-and-so must always happen 8 out of 10 times;" However, there are no useful theorems like this. If there were then probability wouldn't be a good model for uncertainty. The useful theorems of probability have conclusions about probabilities, not conclusions about events actually happening. They have the form "if the probability of ...is such-and-such then the probability of ...(something else) is so-and-so".

11. Nov 15, 2014

### Soumalya

Thanks Tashi!!!!!
I am now far more comfortable with my understanding.:)