Answer to Question #327813 in Statistics and Probability for Val

Question

a farmer produces seeds in pckets for sale. the provbility that seed selected at random will grow is 4/5. if 20 seeds are sown what is the probability that

(A) less than 2 will grow

(B) Less than 2 will not grow

(C) Exactly half the seed will grow

(d) exactly 2 will grow

Answer 1

We have a Bernoulli trial - exactly two possible outcomes, "success" (the seed will grow) and "failure" (the seed will not grow) and the probability of success is the same every time the experiment is conducted (a seed is sown), $p=\cfrac{4}{5=0.8}, q=1-p=1-0.8=0.2.$

The probability of each result:

$P(N=k)=\begin{pmatrix} n \\ k \end{pmatrix}\cdot p^k \cdot q^{n-k}.$

$(A)\ \ P(N<2)=P(X=0)+P(X=1)=\\ =\begin{pmatrix} 20 \\ 0 \end{pmatrix}\cdot 0.8^0 \cdot 0.2^{20}+\begin{pmatrix} 20 \\ 1 \end{pmatrix}\cdot 0.8^1 \cdot 0.2^{19}=\\ =\cfrac{20!}{0!\cdot20!}\cdot1\cdot0.2^{20}+\cfrac{20!}{1!\cdot19!}\cdot0.8\cdot0.2^{19}=\\ =8.49\cdot10^{-13} \approx0.$

$(B)\ \ P(N\ge2)=1-P(X<2)=\\ =1-8.49\cdot10^{-13} \approx1.$

$(C)\ \ P(N=10)=\begin{pmatrix} 20 \\10 \end{pmatrix}\cdot 0.8^{10 }\cdot 0.2^{10}=\\ =\cfrac{20!}{10!\cdot10!}\cdot0.8^{10}\cdot0.2^{10}=0.002.$

$(D)\ \ P(N=2)=\\ =\begin{pmatrix} 20 \\ 2 \end{pmatrix}\cdot 0.8^2 \cdot 0.2^{18}=\\ =\cfrac{20!}{2!\cdot18!}\cdot0.8^2\cdot0.2^{18}=\\ =3.19\cdot10^{-11} \approx0.$