Naked Science Forum
General Science => General Science => Topic started by: vhfpmr on 20/01/2016 12:37:14

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?

"Once is happenstance, twice is coincidence, three times is enemy action" (James Bond)

I have to agree with Alan, 0.1% sounds like enemy action!
Is this a homework question?

There are two possible outcomes: Lost & Not Lost. That makes it a binomial probability.
See: http://en.wikipedia.org/wiki/Binomial_distribution

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.

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Now, if the question were about a mass mailout of 10,000 letters, this approximation is no longer valid, since you expect about 10 to get lost.

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.

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.