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?

Correct answer:

a =  0.867
b =  0

Step-by-step explanation:

n=20 ρ=1/10=110=0.1 1/m  p0=(n0)ρ0(1ρ)n0=(200)0.10(10.1)2000.1216 p1=(n1)ρ1(1ρ)n1=(201)0.11(10.1)2010.2702 p2=(n2)ρ2(1ρ)n2=(202)0.12(10.1)2020.2852 p3=(n3)ρ3(1ρ)n3=(203)0.13(10.1)2030.1901  a=p0+p1+p2+p3=0.1216+0.2702+0.2852+0.19010.867=0.867  p=1a=10.8670.133

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