Multiplication theorem and Independent Probability

Independent Events (Theory)

Two events are said to be an event if happening of one event doesn't affect the other one.

let two event A and B are  independent events then:

P(A ∩ B) = P(A). P(B)

Prove:

from conditional probability theorem, we already know that

P(A/B)= P(A  ∩ B) / P(B) .....................(1)

Since Events are independent then P(E/F)= P(E)

therefore P(A/B)= P(A) because happening of A is not depending on happening of B , they are Independent to each other.

Replacing P(A/B) with P(A) in equation (1)

P(A) . P(B) = P(A ∩ B)

for any number of events if this definition holds we can say that they are Independent events.

Please remember one thing Independent events are not Mutually exclusive events because In mutually exclusive events, A ∩ B = Ø

So P(A ∩ B  ) = 0

 

Contributor's Info

Created:
0Comment
Independent Events Example (1) :Marriage problem

Ambikesh wants to marry his girlfriend and his girlfriend has already told him that she will marry him only when each and every member of her family will like Ambikesh. Each member of his girlfriend's family thinks independently. There are four members of her family.

P(father likes Ambikesh) = 1/3

P(Mother likes Ambikesh) = 3/4

P( Brother likes Ambikesh ) = 1/6

P(Dog likes Ambikesh) = 2/5

What is the probability that  Ambikesh can marry his girlfriend?

 

This question is directly based on the formula of Independent events.

P(marriage)= (1/3) * (3/4) * (1/6)*(2/5)

                        =  1/60 

0Comment
Independece (Probability) | MIT Assignment

You have a fair five-sided die. The sides of the die are numbered from 1 to 5. Each die roll is independent of all others, and all faces are equally likely to come out on top when the die is rolled. Suppose you roll the die twice.

(A)

Let event A to be “the total of two rolls is 10”, event B be “at least one roll resulted in 5”, and event C be “at least one roll resulted in 1”.

  1. Is event A independent of event B?
  2.  Is event A independent of event C?

(B)

Let event D be “the total of two rolls is 7”, event E be “the difference between the two roll outcomes is exactly 1”, and event F be “the second roll resulted in a higher number than the first roll”.

  1. Are events E and F independent?
  2. Are events E and F independent given event D?

 

(A)

(1) NO

Consider the sample space(visualized below) when the five-sided die is rolled twice :

Overall, there are 25 different outcomes in the sample space. For a total of 10, we should get a 5 on both rolls.

Therefore A ⊂ B, and

P(B|A) = \(\frac{P(A \cap B)}{P(A)} = \frac{ P(A)}{P(A)} = 1\)

We observe that to get at least one 5 showing, we can have 5 on the first roll, 5 on the second roll, or 5 on both rolls, which corresponds to 9 distinct outcomes in the sample space. Therefore

\(P(B) = \frac{9}{25} \neq P(B | A)\)

Hence, A and B are not independent.

Ans NO.

(2) NO

Given event A, we know that both roll outcomes must be 5. Therefore, we could not have event C occur, which would require at least one 1 showing. Formally, there are 9 outcomes in C, and

\(P(C) = \frac{9}{25} \)

but, \(P(C | A) = 0 \neq P(C)\)

Hence, Ans NO.

 

(B)

(1) NO

Out of the total 25 outcomes, 5 outcomes correspond to equal numbers in the two rolls. In half of the remaining 20 outcomes, the second number is higher than the first one. In the other half, the first number is higher than the second. Therefore,

\(P(F) = \frac{10}{25}\)

There are eight outcomes that belong to event E:

E = {(1, 2),(2, 3),(3, 4),(4, 5),(2, 1),(3, 2),(4, 3),(5, 4)}.

To find P(F|E), we need to compute the proportion of outcomes in E for which the second number is higher than the first one:

\(P(F | E) = \frac{1}{2} \neq P(F) \)

Hence, Ans is NO.

 

(2) YES,

Conditioning on event D reduces the sample space to just four outcomes

{(2, 5),(3, 4),(4, 3),(5, 2)}

which are all equally likely. It is easy to see that

\(P(E|D) = \frac{2}{4} = \frac{1}{2} \\ P(F | D) = \frac{2}{4} = \frac{1}{2} \\ P(E \cap F | D) = \frac{1}{4} = P(E|D)P(F|D)\)

hence, Ans YES.

0Comment
Some Properties of Independent Events

If A and B are Independent events then:

1.  A and Bc are also independent.

2.  Ac and B  are also independent.

3.  Ac and Bc are also independents.

 

I will prove one and you please try to prove the remaining on your own.

let's prove:

3. Ac and Bc are also independents.

if  Ac and Bc are also independents then 

P(A ∩ Bc) = P(Ac).P(Bc)

P(A ∩ Bc) ¬ P(A ∪ B ) = 1 - P(A ∪ B )

                        =  1 - {P(A) + P(B) - P(A ∩ B) }

                        =  1-   {P(A) + P(B) - P(A). P(B) }

                        =  {1- P(A) } -  P(B)  {1- P(A) }

                        = {1 - P(A)}   {1- P(B)}

                        =  P(Ac).P(Bc)

 

Hence Proved.

   

Contributor's Info

Created:
0Comment
Section 3.4 Example 4f Sheldon Ross

An infinite sequence of independent trials is to be performed. Each trial results in a success with probability p and a failure with probability 1 − p. What is the probability that 
(a) at least 1 success occurs in the first n trials;
(b) exactly k successes occur in the first n trials;
(c) all trials result in successes?

In order to determine the probability of at least 1 success in the first n trials, it is easiest to compute first the probability of the complementary event: that of no successes in the first n trials. If we let Ei denote the event of a failure on the ith trial, then the probability of no successes is, by independence,

P(E1E2 · · · En) = P(E1)P(E2) · · · P(En) = \((1 - p)^n\)

Hence, the answer to part (a) is \(1 - (1 - p)^n\).

(b) Consider any particular sequence of the first \(n\) outcomes containing \(k\) successes and \(n-k\) failures. Each one of these sequences
will, by the assumed independence of trials, occur with probability \(p^k(1-p)^{n-k}\)

There are \(\binom{n}{k}\) such sequences , the desired probability in part (b) is 

\(P\{exactly\ \ k \ \ success\} = \binom{n}{k} p^k(1-p)^{n - k}\)

(c)

The probability of the first n trials all resulting in success is given by

\(P(E_1^cE_2^c....E_n^c) = p^n\)

0Comment
Conditional Probability
Content covered: 

Introduction to Contional Probability.
1. A family has two children. What is the probability that both the children are girls given that at least one of them is girl?
2. One card is drawn randomly from ten cards numbered 1 to 10. If it is known that the number on the drawn card is more then 3, what is the probability that it is an even number?
3. David has a question bank consisting of 300 easy type X, 200 difficult type X, 500 easy type Y and 400 difficult type Y questions. If a auestion is selected at random from the question bank, what is the probability that it will be an easy question given that it is a type Y question?

More Less
1Comment

  • This quiz contains 5 questions on the topic Probability - I
  • Lean well before you attempt the quiz
  • You can attempt the quiz unlimited number of times.

Difficulty Level:  intermediate