91Ó°ÊÓ

The binomial distribution is a probability distribution that represents the chance of a value taking one of two independent values.

a)
Following is a theorem we are using :

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{\rm{ }}\end{array}} \right.\)

(a):

Using the theorem given above, for\(n = 8,x = 3\), and\(p = 0.35\), the following is true

\(\begin{aligned}b(3;8,0.35) &= \left( {\begin{array}{*{20}{l}}8\\3\end{array}} \right){0.35^3}{(1 - 0.35)^{8 - 3}}\\ &= \left( {\begin{array}{*{20}{l}}8\\3\end{array}} \right){0.35^3} \cdot {0.65^5}\\ &= 0.2786.\end{aligned}\)

Therefore, the required binomial property is \(0.2786\).

Using the theorem given above, for\(n = 8,x = 5\), and\(p = 0.6\), the following is true

\(\begin{aligned}b(5;8,0.6) = \left( {\begin{array}{*{20}{l}}8\\5\end{array}} \right){0.6^5}{(1 - 0.6)^{8 - 5}}\\&= \left( {\begin{array}{*{20}{l}}8\\5\end{array}} \right){0.6^5} \cdot {0.4^3}\\ &= 0.2787.\end{aligned}\)

Therefore, the required binomial property is \(0.2787\).

c)
For \(n = 7\), and \(p = 0.6\), the following is true

\(\begin{aligned}P(3 \le X \le 5)\mathop = \limits^{(1)} b(3;7,0.6) + b(4;7,0.6) + b(5;7,0.6)\\ &= \left( {\begin{array}{*{20}{l}}7\\3\end{array}} \right){0.6^3}{(1 - 0.6)^{7 - 3}}\\ &+ \left( {\begin{array}{*{20}{l}}7\\4\end{array}} \right){0.6^4}{(1 - 0.6)^{7 - 4}}\\ &+ \left( {\begin{array}{*{20}{l}}7\\5\end{array}} \right){0.6^5}{(1 - 0.6)^{7 - 5}}\\ &= 0.1935 + 0.2903 + 0.2613\\ &= 0.7451\end{aligned}\)

Therefore, from 3 to 5 , binomial random variable X can take only values 3,4,5.

d)

For\(n = 9\), and\(p = 0.1\), the following is true

\(\begin{aligned}P(1 \le X)\mathop = \limits^{(1)} 1 - P(X = 0)\\ &= 1 - \left( {\begin{array}{*{20}{l}}9\\0\end{array}} \right){0.1^3}{(1 - 0.1)^{9 - 0}}\\ &= 1 - 0.3874\\ &=0.6126\end{aligned}\)
Therefore, the complement of event\(\{ 1 \le X\} \)is\(\{ X = 0\} \).