What Are the Chances of Water Rationing Over a Decade in Southland?

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Water rationing in Southland has a 15% probability each year, leading to an expected 1.5 years of rationing over a decade. The probability of no water rationing in the next ten years is calculated as approximately 19.7%. To find the likelihood of at least two years of rationing, one must consider the complementary probabilities of zero and one year of rationing, which total about 83%. The correct probability for at least two years of rationing is therefore about 17%. Understanding these calculations is crucial for effective water management policies in the region.
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Water shortages require water rationing policies. From past records, we know that the probability that water needs to be rationed in a southland water district in any given year is 0.15. Assume that water rationing in consecutive years are independent events.

Let X be the number of years with water rationing in a sample of ten years. What is the expected (mean) number of years with water rationing in 10 years for a southland water district?
For this one E(X)=np=1.5 i think

What is the probability that there would be no water rationing in the next ten years?

I'm not sure how to calculate this...i know at least for one year it should be .85 is that right?

What is the probability that there would be water rationing in at least two of the next ten years?
I'm assuming this would be 2*.15
 
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vipertongn said:
Water shortages require water rationing policies. From past records, we know that the probability that water needs to be rationed in a southland water district in any given year is 0.15. Assume that water rationing in consecutive years are independent events.

Let X be the number of years with water rationing in a sample of ten years. What is the expected (mean) number of years with water rationing in 10 years for a southland water district?
For this one E(X)=np=1.5 i think

Yes. Do you recognize a binomial distribution here?

What is the probability that there would be no water rationing in the next ten years?

I'm not sure how to calculate this...i know at least for one year it should be .85 is that right?

Yes.

What is the probability that there would be water rationing in at least two of the next ten years?
I'm assuming this would be 2*.15

No. What is the probability for (A AND B) if A and B are independent?
 
I hate to bring up an old thread, but I'm reviewing this and i haven't gotten the answer for the last two really.
What is the probability that there would be no water rationing in the next ten years?
ok, I think for this case I calculate it first the probability of it actually not occurring, which is...
P(X=0)= 10C0*p^0(0.85)^10= 0.197

Which I hope is correct...

Then for the next one I calculate the probabilities of it occurring at LEAST 2 times by finding the probabilities of 1 and 2 happening and then adding them with the .197
P(X=1)=.35
P(X=2)=.27
so it would become about .83. For it to at least be occurring 1-.83 = .17 of 2 rationings occurring

I hope I'm correct
 
vipertongn said:
I hate to bring up an old thread, but I'm reviewing this and i haven't gotten the answer for the last two really.
What is the probability that there would be no water rationing in the next ten years?
ok, I think for this case I calculate it first the probability of it actually not occurring, which is...
P(X=0)= 10C0*p^0(0.85)^10= 0.197

Which I hope is correct...

Yes.

Then for the next one I calculate the probabilities of it occurring at LEAST 2 times by finding the probabilities of 1 and 2 happening and then adding them with the .197
P(X=1)=.35
P(X=2)=.27
so it would become about .83. For it to at least be occurring 1-.83 = .17 of 2 rationings occurring

I hope I'm correct

No. The complementary event to "at least two" is "zero or one". You just need to subtract the probabilities for 0 and 1.
 
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