Solving Word Problems Involving Binomial Probability Distribution


Solving Problems Involving Binomial Probability Distribution



If an experiment can result in two outcomes, success and failure, with probabilities p and q, respectively then the probability distribution of X, the number of successes in an independent trial is given by :


P ( X = x ) = b (X; n, p) = nCx p^x q^n-x

Where X = 0, 1, 2, 3,……….n


Properties of Binomial Probability Distribution

1) The experiment consists of n repeated independent trials.

2) Each trial has two possible outcomes, success and failure.

3) The probability of success remain constant from trial to trial.



Sample Problem Number One :

Assume that each child has probability of 0.5 of being a boy. Find the probability that a family of five children has :

a) Exactly one boy

b) At least has two boys

c) At most three boys



Solution :

Given :

p = probability that a child is a boy = 0.5

q = 1 – p = 1 – 0.5 = 0.5

n = 5

(a) Exactly one boy

P(X = 1) = b ( 1;5,0.5) = 5C1 (0.5)^1(0.5)^4

5C1 = 5!/(5-1)! 1! = 5!/4!1! = 5/1 = 5

P(X = 1) = 5 (0.5)(0.0625) = 0.15625



b) At least 4 boys

P(X > = 4) = P(X = 4 ) + P(X = 5 )

= b (4; 5, 0.5) + b (5; 5, 0.5)

=5C4(0.5)^4 (0.5)^1 + 5C5 (0.5)^5 (0.5)^0

5C4 = 5!/(5-4)! 4! = 5/1 = 5 5C5 = 5!/(5-5)! 5! = 1

= 5(0.5)^4(0.5) + (1) (0.5)^5 (1)

= 5 (0.0625)(0.5) + (1)(0.03125) (1)

= 0.15625 + 0.03125

=0.1875


c) At most three boys

P(X < = 3) = P(X= 0) + P(X = 1) + P(X = 2) + P(X = 3)

= b(0: 5, 0.5) + b(1; 5, 0.5) + b(2; 5,0.5) + b(3; 5, 0.5)

=5C0 (0.5)^0 (0.5)^5 + 5C1(0.5)^1 (0.5)^4 + 5C2 (0.5)^2 (0.5)^3 + 5C3 (0.5)^3(0.5)^2

5C0 = 5!/(5-0)!0! = 1

5C1 = 5!/(5-1)!1! = 5!/4! = 5

5C2 = 5!/(5-2)!2! = 5!/3! 2! = 10

5C3 = 5!/(5-3)! 3! = 5!/2! 3! = 10

P(X = 0 ) = (1)(1)(0.5)^5 = 0.03125

P(X= 1 ) = (5 ) (0.5) (0.0625) = 0.15625

P(X = 2 ) = (10) (0.25) (0.125) = 0.3125

P(X = 3 ) = (10) (0.125) (0.25) = 0.3125

P( X <= 3 ) = 0.03125 + 0.15625 + 0.3125 + 0.3125 = 0.8125



Sample Problem Number Two

The probability that a certain kind of component will survive a given shock is ¾. Find the probability that exactly 2 of the next 4 components tested survive.


Solution

Given :

p = 0.75

q = 1 – 0.75 = 0.25

n = 4

Let X = number of components

P(X = 2) = b( 2; 4 , 0.75) = 4C2 (0.75)^2 (0.25)^2

4C2 = 4!/(4-2)! 2! = 4!/2! 2! = 12/2 = 6

P(X = 2) = 6 (0.75)^2 (0.25)^2

= 6 (0.5625)(0.0625 )

=0.2109375




SOURCE :


BASIC STATISTICS By

Ang

Billones

Dechavez

Diansuy














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Comments 10 comments

Sinea Pies profile image

Sinea Pies 4 years ago from Northeastern United States

Well, you've lost me but I am very impressed with your mathematical mind. Somebody will love AND understand this hub and will use it often.


Eddie-Perkins 4 years ago

I was an A+ student in math Maria Cristina, but that has been a few years. Now I have to agree with Sinea; this hurts my brain – at least what is left of it. Vote up interesting and useful (to someone:). God bless you sister. ~ eddie


Fraser Soul profile image

Fraser Soul 4 years ago from Bloomfield, CT

Math was never my best subject. I'm very impressed and I agree with Eddie it hurts my brian at least what left of it. God bless you and me too for surving the review.


cristina327 profile image

cristina327 4 years ago from Manila Author

Hi Sinea Pies it is great to hear from you. Thank you for taking time to read this hub. Your visit and comments are much appreciated. Remain blessed always. Best regards.


cristina327 profile image

cristina327 4 years ago from Manila Author

Hi Eddie Perkins it is good to see you here in this Math hub. I am glad to know that you used to be an A+ student in Math. I am glad you find this hub interesting. Thank you for dropping by and appreciating this hub and even for voting for it as useful and interesting. May you be blessed today and always. Best regards.


cristina327 profile image

cristina327 4 years ago from Manila Author

Hi Fraser Soul it is great to hear from you. Thank you for gracing and appreciating this hub. Your visit is much appreciated. Blessings to you always. Best regards.


HOOWANTSTONO profile image

HOOWANTSTONO 4 years ago

I last did maths at school, many years ago, but maths still doesnt change .

Good show.


cristina327 profile image

cristina327 4 years ago from Manila Author

Hi HOOWANTSTONO glad to hear from you again. Thank you for gracing this hub. Your visit is much appreciated. Blessings to you and regards.


DeBorrah K. Ogans profile image

DeBorrah K. Ogans 4 years ago

Cristina, You are not only cognizant spiritually but a mathematical whiz as well! Thank you for sharing, In His Love, Peace & Blessings!


cristina327 profile image

cristina327 4 years ago from Manila Author

Hi DeBorrah it is good to see you here in this mathematical hub.Thank you for gracing this hub. Your visit is always a blessing. Remain blessed always and best regards.

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