Twenty

Twenty swallows sit on a 10 m long telephone cable. Assume that swallows are completely randomly distributed along the line.

(a) What is the probability that more than three swallows sit on a randomly selected section of cable 1 m long?

(b) What is the probability that the spacing between two swallows will be greater than 1 m and, at the same time, less than 2 m?

Final Answer:

a =  0.867
b =  

Step-by-step explanation:

$$\begin{gathered} n=20 \\ \rho =1/10=\frac{1}{10}=0.11\text{/m} \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot {\rho }^0\cdot (1-\rho {)}^{n-0}=(\genfrac{}{}{0ex}{}{20}{0})\cdot 0.1^0\cdot (1-0.1{)}^{20-0}=1\cdot 0.1^0\cdot 0.9^{20}\doteq 0.1216 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot {\rho }^1\cdot (1-\rho {)}^{n-1}=(\genfrac{}{}{0ex}{}{20}{1})\cdot 0.1^1\cdot (1-0.1{)}^{20-1}=20\cdot 0.1^1\cdot 0.9^{19}\doteq 0.2702 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot {\rho }^2\cdot (1-\rho {)}^{n-2}=(\genfrac{}{}{0ex}{}{20}{2})\cdot 0.1^2\cdot (1-0.1{)}^{20-2}=190\cdot 0.1^2\cdot 0.9^{18}\doteq 0.2852 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot {\rho }^3\cdot (1-\rho {)}^{n-3}=(\genfrac{}{}{0ex}{}{20}{3})\cdot 0.1^3\cdot (1-0.1{)}^{20-3}=1140\cdot 0.1^3\cdot 0.9^{17}\doteq 0.1901 \\ a=p_0+p_1+p_2+p_3=0.1216+0.2702+0.2852+0.1901\doteq 0.867 \\ p=1-a=1-0.867\doteq 0.133 \end{gathered}$$

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