# 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**

## Comments

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.

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

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.

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

Good show.

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

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