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

Answer:

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

Step-by-step explanation:

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

$(a + b)^{n} = \Sigma \limits^{n}_{k = 0}\,\frac{n!}{k!\cdot (n-k)!}\cdot a^{n-k}\cdot b^{k}$ (1)

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

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

$(0.3 + q)^{5} = \frac{243}{100000} + \frac{81}{2000}\cdot q + \frac{27}{100}\cdot q^{2} + \frac{9}{10} \cdot q^{3} + \frac{3}{2}\cdot q^{4} + q^{5}$