Question #267597

The number of injury claims per month is modeled by a random variable 𝑁 with Pr(𝑁 = 𝑛) = 1/(𝑛+1)(𝑛+2), for 𝑛 = 0, 1, 2, 3, · · · . Determine the probability of at least one claim during a particular month, given that there have been at most four claims during the months

1
Expert's answer
2021-11-17T18:03:10-0500

We need to compute P(N1N4),P(N ≥ 1|N ≤ 4),  which is the same as P(1N4)P(N4)\dfrac{P(1\leq N\leq 4)}{P(N\leq 4)}

P(1N4)=P(N=1)+P(N=2)P(1\leq N\leq 4)=P(N=1)+P(N=2)

+P(N=3)+P(N=4)+P(N=3)+P(N=4)

=1(1+1)(1+2)+1(2+1)(2+2)=\dfrac{1}{(1+1)(1+2)}+\dfrac{1}{(2+1)(2+2)}

+1(3+1)(3+2)+1(4+1)(4+2)+\dfrac{1}{(3+1)(3+2)}+\dfrac{1}{(4+1)(4+2)}

=16+112+120+130=13=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{3}

P(N4)=P(N=0)+P(1N4)P( N\leq 4)=P(N=0)+P(1\leq N\leq 4)

=1(0+1)(0+2)+13=56=\dfrac{1}{(0+1)(0+2)}+\dfrac{1}{3}=\dfrac{5}{6}

Then


P(N1N4)=P(1N4)P(N4)P(N ≥ 1|N ≤ 4)=\dfrac{P(1\leq N\leq 4)}{P(N\leq 4)}

=1/35/6=25=0.4=\dfrac{1/3}{5/6}=\dfrac{2}{5}=0.4


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS