Question

Use the binomial expression (p+q)^n to calculate a binomial distribution with n=5 and p=0.3.(Show all steps)

Answer 1

The binomial in expanded form is
`(0.3 + q)^(5) = (243)/(100000) + (81)/(2000)\cdot q + (27)/(100)\cdot q^(2) + (9)/(10) \cdot q^(3) + (3)/(2)\cdot q^(4) + q^(5)`.

Explanation:

The Binomial Theorem states that a binomial of the form
`(a + b)^(n)` can be expanded by using the following identity:

`(a + b)^(n) = \Sigma \limits^(n)_(k = 0)\,(n!)/(k!\cdot (n-k)!)\cdot a^(n-k)\cdot b^(k)` (1)

If we know that
`a = p = 0.3` and
`n = 5`, then the expanded form of the binomial is:

`(p+q)^(n) = (243)/(100000) + 5\cdot \left((81)/(10000) \right)\cdot q + 10\cdot \left((27)/(1000))\cdot q^(2) + 10\cdot \left((9)/(100) \right)\cdot q^(3) + 5\cdot \left((3)/(10) \right)\cdot q^(4) + q^(5)`

`(0.3 + q)^(5) = (243)/(100000) + (81)/(2000)\cdot q + (27)/(100)\cdot q^(2) + (9)/(10) \cdot q^(3) + (3)/(2)\cdot q^(4) + q^(5)`