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=101=0.1 1/m  p0=(0n)ρ0(1ρ)n0=(020)0.10(10.1)200=10.100.9200.1216 p1=(1n)ρ1(1ρ)n1=(120)0.11(10.1)201=200.110.9190.2702 p2=(2n)ρ2(1ρ)n2=(220)0.12(10.1)202=1900.120.9180.2852 p3=(3n)ρ3(1ρ)n3=(320)0.13(10.1)203=11400.130.9170.1901  a=p0+p1+p2+p3=0.1216+0.2702+0.2852+0.19010.867  p=1a=10.8670.133

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