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