How do you solve probability questions that involve the phrase "at least"?

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Offline vhfpmr

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If the probability of a letter getting lost in the post is 0.1%, how do I calculate the probability of 5 letters getting lost out of a total of 9, and the probability of at least 5 getting lost?
« Last Edit: 21/01/2016 11:23:25 by chris »

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Offline alancalverd

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Re: Probability Question
« Reply #1 on: 20/01/2016 16:37:57 »
"Once is happenstance, twice is coincidence, three times is enemy action" (James Bond)
helping to stem the tide of ignorance

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Offline Colin2B

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Re: Probability Question
« Reply #2 on: 21/01/2016 00:22:55 »
I have to agree with Alan, 0.1% sounds like enemy action!

Is this a homework question?
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline evan_au

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Re: Probability Question
« Reply #3 on: 21/01/2016 09:12:01 »
There are two possible outcomes: Lost & Not Lost. That makes it a binomial probability.

See: http://en.wikipedia.org/wiki/Binomial_distribution

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Offline Colin2B

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You can use the binomial to calc 5 lost and cumulative to calc at least 5.
If you want to see the raw mechanism, and if this is not a homework question, i'd be happy to post an explanation.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline evan_au

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Quote
that involve the phrase "at least 5"?
You can calculate the probability that 6 get lost, and repeat for 7, 8 & 9. Then add them up. This will give you an exact answer, but it becomes laborious if the question is about a mass mailout of 1,000 letters.

Or you can take a shortcut, and observe that because the probability that an individual letter gets lost is pretty low (0.1%), the probability that most of the letters will get lost is extremely low.

So you can approximate the probability of "more than 5 getting lost" as "the probability of 6 getting lost", knowing that this will be a slight underestimate.

Now, if the question were about a mass mailout of 10,000 letters, this approximation is no longer valid, since you expect about 10 to get lost.

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Offline mrsmith2211

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Each letter has a .001 chance of getting lost. So .001/5 I calculate it out to a .0002 chance of 5 getting lost if 5 letters sent, then 5/9=.55555... and .0002 x .55555 =.000111 or .011%, Open to being wrong.
« Last Edit: 23/01/2016 03:13:56 by mrsmith2211 »

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Offline alysdexia

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Each letter has a .001 chance of getting lost. So .001/5 I calculate it out to a .0002 chance of 5 getting lost if 5 letters sent, then 5/9=.55555... and .0002 x .55555 =.000111 or .011%, Open to being wrong.

Multiply; don't add.