June 19 , 2012 - Lesson on Binomial Distribution

       Everyday subjects like that of the Advance Statistics, give me a lot of things to worry. One thing is about the assignment-will my assignment be able to have a passing score? Next is about the lesson if I am going to  be able to absorb it well or if it will be another time wasted because of  nothing learned due to my very slow mind. And to worry if I can get out of the room with my correct passport (thanks for my other classmates for helping me got the correct answer) :). Stat is a good subject if  you are good at it,  especially if you are great in understanding word problems,  memorizing formulas, and  being good in solving through your calculator.
  
     Well , for Tuesday (June 19, 2012) our lesson is all about  Binomial Distribution. I learned that a binomial experiment is a probability experiment that satisfies 4 requirements (trials that have two outcomes or outcomes that can be reduced to two outcomes;  there must  be a fixed number of trials; the outcomes of each trial must be  independent of each other and the probability of success must remain the same for each trial).
 
       I also had recalled the Binomial Probability Formula:

 P(x ) =          n!              . p^x.q^{n - x}        
                 ( n-X )!  X!              

 We also have learned about Mean, Variance and Standard deviation for Binomial Distribution; and the Multinomial Distribution, and the following are the formulas:

  Mean            
                   μ = n   x   p
  ^Variance           

                   σ²= n*p*q        

  Standard Deviation               

                   σ= square root of n*p *q      

  

   Multinomial Distribution 

                           P = [ n! / ( x₁! *  x₂! * ... x_n! ) ] * ( p₁^x₁ * p₂^x₂ * . . . * p_k^x_k )

          

       

     It is an achievement for my day if I had learned much in school especially to subjects that I am weak, like Advance Statistics.


 by: Kent Spencer  Manalo Mendez

 







Examples on Binomial Probability Distribution


Example 1

What is the probability of obtaining 45 or fewer heads in 100 tosses of a coin?
Solution: To solve this problem, we compute 46 individual probabilities, using the binomial formula. The sum of all these probabilities is the answer we seek. Thus,

b(x < 45; 100, 0.5) = b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + . . . + b(x = 45; 100, 0.5)
b(x < 45; 100, 0.5) = 0.184

Example 2

The probability that a student is accepted to a prestigious college is 0.3. If 5 students from the same school apply, what is the probability that at most 2 are accepted?
Solution: To solve this problem, we compute 3 individual probabilities, using the binomial formula. The sum of all these probabilities is the answer we seek. Thus,

b(x < 2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5, 0.3)
b(x < 2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087
b(x < 2; 5, 0.3) = 0.8369

Example 3

What is the probability that the world series will last 4 games? 5 games? 6 games? 7 games? Assume that the teams are evenly matched.
Solution: This is a very tricky application of the binomial distribution. If you can follow the logic of this solution, you have a good understanding of the material covered in the tutorial, to this point.
In the world series, there are two baseball teams. The series ends when the winning team wins 4 games. Therefore, we define a success as a win by the team that ultimately becomes the world series champion.
For the purpose of this analysis, we assume that the teams are evenly matched. Therefore, the probability that a particular team wins a particular game is 0.5.
Let's look first at the simplest case. What is the probability that the series lasts only 4 games. This can occur if one team wins the first 4 games. The probability of the National League team winning 4 games in a row is:

b(4; 4, 0.5) = ₄C₄ * (0.5)⁴ * (0.5)⁰ = 0.0625

Similarly, when we compute the probability of the American League team winning 4 games in a row, we find that it is also 0.0625. Therefore, probability that the series ends in four games would be 0.0625 + 0.0625 = 0.125; since the series would end if either the American or National League team won 4 games in a row.
Now let's tackle the question of finding probability that the world series ends in 5 games. The trick in finding this solution is to recognize that the series can only end in 5 games, if one team has won 3 out of the first 4 games. So let's first find the probability that the American League team wins exactly 3 of the first 4 games.

b(3; 4, 0.5) = ₄C₃ * (0.5)³ * (0.5)¹ = 0.25

Okay, here comes some more tricky stuff, so listen up. Given that the American League team has won 3 of the first 4 games, the American League team has a 50/50 chance of winning the fifth game to end the series. Therefore, the probability of the American League team winning the series in 5 games is 0.25 * 0.50 = 0.125. Since the National League team could also win the series in 5 games, the probability that the series ends in 5 games would be 0.125 + 0.125 = 0.25.
The rest of the problem would be solved in the same way. You should find that the probability of the series ending in 6 games is 0.3125; and the probability of the series ending in 7 games is also 0.3125.

    

From: stattrek.com/...distributions/binomial.aspx

By : Kent Spencer Manalo Mendez