# Compute the following binomial probabilities directly from the formula for b(x, n, p).

1. b( 3, 8, 0.6 )
2. b( 5, 8, 0.6 )
3. P( 3 $\le$ X $\le$ 5 ) when n = 8 and p = 0.6

The aim of this question is to use the binomial random variable and its probability mass function to find probability values.

The binomial probability mass function is mathematically defined as:

$P( \ X \ = \ x \ ) \ = \ b( \ x, \ n, \ p \ ) \ = \ \left ( \begin{array}{c} n \\ x \end{array} \right ) \ p^x \ ( \ 1 \ – \ p \ )^{ n – x }$

Part (a) – b( 3, 8, 0.6 )

$b( \ 3, \ 8, \ 0.6 \ ) \ = \ \left ( \begin{array}{c} 8 \\ 3 \end{array} \right ) \ (0.6)^3 \ ( \ 1 \ – \ 0.6 \ )^{ 8 – 3 }$

$b( \ 3, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 3! \ (8 – 3)! } \ (0.6)^3 \ ( \ 0.4 \ )^5$

$b( \ 3, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 3! \ 5! } \ (0.6)^3 \ (0.4)^5$

$b( \ 5, \ 8, \ 0.6 \ ) \ = \ (56) \ (0.6)^3 \ (0.4)^5$

$b( \ 3, \ 8, \ 0.6 \ ) \ = \ 0.1238$

– b( 5, 8, 0.6 )

$b( \ 5, \ 8, \ 0.6 \ ) \ = \ \left ( \begin{array}{c} 8 \\ 5 \end{array} \right ) \ (0.6)^5 \ ( \ 1 \ – \ 0.6 \ )^{ 8 – 5 }$

$b( \ 5, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 5! \ (8 – 5)! } \ (0.6)^5 \ ( \ 0.4 \ )^3$

$b( \ 5, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 5! \ 3! } \ (0.6)^3 \ (0.4)^5$

$b( \ 5, \ 8, \ 0.6 \ ) \ = \ (56) \ (0.6)^5 \ (0.4)^3$

$b( \ 5, \ 8, \ 0.6 \ ) \ = \ 0.2787$

– P( 3 $\le$ X $\le$ 5 ) when n = 8 and p = 0.6

Using same approach as part (a) and (b):

$P( \ X \ = \ 4 \ ) \ = \ b( \ 4, \ 8, \ 0.6 \ ) \ = \ 0.2322$

Since:

$P( \ 3 \le X \le 5 \ ) \ = \ P( \ X \ = \ 3 \ ) \ + \ P( \ X \ = \ 4 \ ) \ + \ P( \ X \ = \ 5 \ )$

$P( \ 3 \le X \le 5 \ ) \ = \ 0.1238 \ + \ 0.2322 \ + \ 0.2787$

## Numerical Result

b( 3, 8, 0.6 ) = 0.1238

b( 5, 8, 0.6 ) = 0.2787

P( 3 $\le$ X $\le$ 5 ) = 0.6347

## Example

Find the probability P( 1 $\le$ X ) where X is a random variable with n = 12 and p = 0.1

Using same approach as part (a) and (b):

$P( \ X \ = \ 0 \ ) \ = \ b( \ 0, \ 12, \ 0.1 \ ) \ = \ 0.2824$

Since:

$P( \ 1 \le X \ ) \ = \ 1 \ – \ P( \ X \le 1 \ ) \ = \ 1 \ – \ P( \ X \ = \ 0 \ )$

$P( \ 1 \le X \ ) \ = \ 1 \ – \ 0.2824 \ = \ 0.7176$