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Q-) The number of injury claims per month is modelled by a random variable N with: P(N=n)= 1/(n+1)(n+2),where n>=0 Determine the probability of at least 1 claim during a particular month, GIVEN that there have been at most 4 claims during that month

A-) I tried punching in values of n. So I got the conditional probability as = (I/6 +1/12+ 1/20+ 1/30)/(1/3 + 1/6 + 1/12 + 1/20 + 1/30) But this isn't matching with the given answer. I am not sure what's wrong. If at all it is wrong. Thanks in advance.

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    $\begingroup$ 'self-study' questions, if this is one, must have the tag saying self-study. See stats.stackexchange.com/tags/self-study/info $\endgroup$ – Carl Jul 5 '18 at 6:11
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    $\begingroup$ (Welcome to the site!) It would be also helpful to give a more detailed explanation of what you tried (which conditional probability you are evaluating, where do the numbers come from etc.) as well as the final answer vs. the given answer. While you edit the tag and the answer, please also change the tag condition-number (presumably you meant conditional probability, condition numbers are something different) $\endgroup$ – Juho Kokkala Jul 5 '18 at 6:58
  • $\begingroup$ Where did "1/3" come from in the denominator? $\endgroup$ – whuber Jul 5 '18 at 11:44
  • $\begingroup$ Oh yes..my bad I will keep that in mind @juho kokkala $\endgroup$ – Slowdown Sloth Jul 9 '18 at 15:49
  • $\begingroup$ 1/3 is the value when n=0,@whuber $\endgroup$ – Slowdown Sloth Jul 9 '18 at 15:49
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$$ P(N=n) = \frac{1}{(n+1)(n+2)} $$

$$ P(N \neq 0 | N \leq 4) = 1 - P(N=0 | N \leq 4) = 1- \frac{P(N = 0) \cdot P(N \leq4)}{P(N \leq 4)} $$

$$ P(N \leq 4) = P(N=0, 1, 2, 3, 4) \\ = P(N=0) + P(N=1) + P(N=2) + P(N=3) + P(N=4) \\ $$

You can now solve the rest by yourself, since you know the formula for $P(N=n)$.

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  • $\begingroup$ In the second line you appear to assume the events $N=0$ and $N\le 4$ are independent, but obviously that's not the case. $\endgroup$ – whuber Jul 7 '18 at 1:46
  • $\begingroup$ I apologize, I have not looked at this material in quite a long time. Please offer your advice, I would genuinely be interested in knowing how you would approach the question. $\endgroup$ – ERT Jul 7 '18 at 1:57
  • $\begingroup$ I would apply the formula for conditional probability directly, much as in the question itself. To simplify the answer I would first work out the CDF, which has a very simple form. $\endgroup$ – whuber Jul 7 '18 at 2:02
  • $\begingroup$ Thank you guys.I figured out my mistake. Wrong value of n=0. $\endgroup$ – Slowdown Sloth Jul 9 '18 at 16:01
  • $\begingroup$ But @whuber why do you think it will be a good idea find cdf here? It's a discrete random variable. I will have eventually have to work out and add all the values(and x>=0). Or is there a way to find a generalized cdf for a discrete random variable ? $\endgroup$ – Slowdown Sloth Jul 9 '18 at 16:05

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