Homework Q5

Andrew87 · #1 02-21-2015, 10:34 AM

Hi, I'm working on the 6-th question but I can't get the solution.

I solved Q5 and I found a probability to get $\upsilon = 0$ on just one sample (N=10) of 0.0047. Then to compute the probability to get at least one sample with $\upsilon = 0$ , I modeled the number of samples with 0 red marbles

with a binomial random variable

. Hence, the probability I need is

$P(N_0 \geq 1) = 1 - P(N_0 = 0)$

and

$P(N_0 = 0) = (1 - 0.0047)^{1000} = 0.009$

Finally $P(N_0 \geq 1) = 0.991$ , so my closest answer is [3] = 0.550, however the solution report [c] as right answer. Where did I make a mistake ? Thank you in advance for your time (and sorry for my english).

yaser · #2 02-28-2015, 07:55 PM

Quote:

Originally Posted by Andrew87

Hi, I'm working on the 6-th question but I can't get the solution.

I solved Q5 and I found a probability to get $\upsilon = 0$ on just one sample (N=10) of 0.0047. Then to compute the probability to get at least one sample with $\upsilon = 0$ , I modeled the number of samples with 0 red marbles

with a binomial random variable

. Hence, the probability I need is

$P(N_0 \geq 1) = 1 - P(N_0 = 0)$

and

$P(N_0 = 0) = (1 - 0.0047)^{1000} = 0.009$

Finally $P(N_0 \geq 1) = 0.991$ , so my closest answer is [3] = 0.550, however the solution report [c] as right answer. Where did I make a mistake ? Thank you in advance for your time (and sorry for my english).

I did the math and the probability that one sample has no red marbles, which is $0.45^{10}$ evaluates to approximately 0.00034 rather than 0.0047.

Andrew87 · #3 03-03-2015, 07:25 AM

Thank you very much Prof. Yaser !

I did a mistake in calculating the probability of the Homework 4. This was caused by the fact that I used the Hoeffding inequality to compute this probability instead of a simple binomial distribution. Now, I'm wondering a more general question: is this slack between the exact probability one gets with the binomial (in this case) and the bound provided by the Hoeffding inequality typical ? Thank you very much in advance for your answer.

yaser · #4 03-03-2015, 08:47 PM

Yes. Since the bound is the same for cases that have different probabilities, it is forced to be loose in some of the cases.

Andrew87 · #5 03-04-2015, 04:15 AM

Thank you very much for your answer Prof. Yaser !

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