91Ó°ÊÓ

Denote with\({{\rm{A}}_{\rm{1}}}\)and\({{\rm{A}}_{\rm{2}}}\)events –

\({{\rm{A}}_{\rm{1}}}{\rm{ = }}\){a seed from maize ears carries single spikelet};

\({{\rm{A}}_{\rm{2}}}{\rm{ = }}\){a seed produces an ear with single spikelet}.

The following is given –

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.4}}\\{\rm{P}}\left( {{\rm{A}}_{\rm{1}}'} \right){\rm{ = 0}}{\rm{.6}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.29}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {\rm{A}}_{\rm{1}}'} \right){\rm{ = 0}}{\rm{.26}}\end{array}\)

Ten seeds are randomly selected.

(a)

Let random variable\({\rm{X}}\)represent the number of seeds that carry a single spikelet and produce an ear with a single spikelet out of\({\rm{10}}\)randomly selected seeds. The random variable follows Binomial Distribution –

\({\rm{X\sim Bin(10,P(}}{{\rm{A}}_{\rm{1}}} \cap {{\rm{A}}_2}{\rm{))}}\)

The following holds –

\(\begin{array}{c}{\rm{P(}}{{\rm{A}}_{\rm{1}}} \cap {{\rm{A}}_2}{\rm{)}}\mathop {\rm{ = }}\limits^{(1)} {\rm{P(}}{{\rm{A}}_{\rm{1}}}{\rm{)}} \cdot {\rm{P(}}{{\rm{A}}_{\rm{2}}}{\rm{|}}{{\rm{A}}_{\rm{1}}}{\rm{)}}\\{\rm{ = 0}}{\rm{.4}} \cdot {\rm{0}}{\rm{.29 = 0}}{\rm{.116}}\end{array}\)

\({\rm{(1)}}\): the multiplication rule.

The Multiplication Rule –

\({\rm{P(A}} \cap {\rm{B) = P(A|B)}} \cdot {\rm{P(B)}}\)

Therefore, it is obtained –

\({\rm{X\sim Bin(10,}}0.116{\rm{)}}\)

The\({\rm{pmf b(x;n,p)}}\)of Binomial Distribution is given with the theorem below.

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

The requested probability is –

\(\begin{aligned}P(X = 5) &= \left( {\begin{array}{*{20}{l}}n \\ x \end{array}} \right){p^x}{(1 - p)^{n - x}} \\ &= \left( {\begin{array}{*{20}{c}}{10} \\ 5 \end{array}} \right){0.116^5}{(1 - 0.116)^{10 - 5}} \\ &= 0.00286 \\ \end{aligned} \)

Therefore, the value is obtained as \({\rm{0}}{\rm{.00286}}\).

(b)

Let random variable\({\rm{Y}}\)represent the number of seeds that produces ear with a single spikelet out of\({\rm{10}}\)randomly selected seeds. The random variable follows Binomial Distribution –

\({\rm{Y\sim Bin(10,P(}}{{\rm{A}}_2}{\rm{))}}\)

Find\({\rm{P(}}{{\rm{A}}_2}{\rm{)}}\), the following is true –

\(\begin{array}{c}{\rm{P(}}{{\rm{A}}_2}{\rm{)}}\mathop {\rm{ = }}\limits^{(1)} {\rm{P(}}{{\rm{A}}_{\rm{1}}}{\rm{)}} \cdot {\rm{P(}}{{\rm{A}}_{\rm{2}}}{\rm{|}}{{\rm{A}}_{\rm{1}}}{\rm{) + P(A}}_1'{\rm{)}} \cdot {\rm{P(}}{{\rm{A}}_{\rm{2}}}{\rm{|A}}_1'{\rm{)}}\\{\rm{ = 0}}{\rm{.4}} \cdot {\rm{0}}{\rm{.29 + 0}}{\rm{.6}} \cdot {\rm{0}}{\rm{.26 = 0}}{\rm{.272}}\end{array}\)

\({\rm{(1)}}\): use the law of total probability.

The Law of Total Probability –

If \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{,}}...{{\rm{A}}_{\rm{n}}}\) mutually exclusive, and \( \cup _{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = S}}\), then for any event \({\rm{B}}\) the following is true are –

\(\begin{array}{c}{\rm{P(B) = P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ + \ldots + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{n}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{n}}}} \right)\\{\rm{ = }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)\end{array}\)

Therefore, it is obtained –

\({\rm{Y\sim Bin(10,}}0.272{\rm{)}}\)

The\({\rm{pmf b(x;n,p)}}\)of Binomial Distribution is given with the theorem below.

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

The requested probability is –

\(\begin{aligned}P(Y = 5) &= \left( {\begin{array}{*{20}{l}}n \\ x \end{array}} \right){p^x}{(1 - p)^{n - x}} \\ &= \left( {\begin{array}{*{20}{c}} {10} \\ 5 \end{array}} \right){0.272^5}{(1 - 0.0.272)^{10 - 5}} \\ &= 0.0767 \\ \end{aligned} \)

Cumulative Density Function cdf of binomial random variable\({\rm{X}}\)with parameters\({\rm{n}}\)and\({\rm{p}}\)is –

\({\rm{B(x;n,p) = P(X}} \le {\rm{x) = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p),}}\;\;\;{\rm{x = 0,1, \ldots ,n}}\)

And, for the second part, the probability that at most five ears have single spikelet is –

\(\begin{aligned} P(Y \leqslant 5) &= \sum\limits_{x = 0}^5 {\left( {\begin{array}{*{20}{c}}{10} \\ x \end{array}} \right){{0.272}^x} \cdot } {(1 - 0.0.272)^{10 - x}} \\ &= B(5;10,0.272) \\ &= 0.0970 \\ \end{aligned} \)

Therefore, the value is obtained as\({\rm{0}}{\rm{.0970}}\).