Average number
If the probability of an event happening in one trial is p, the average number of trials needed to get the event is 1/p. Why is that true?
Labels: Probability
posted by Chris at
12:24 PM
Thursday, January 28, 2010
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5 Comments:
"If the probability of an event happening in one trial is p..."
Ok, let's say the probability is 1/4. or 25%, p = 1/4 then
.. the average number of trials needed to get the event is 1/p. Why is that true?
That says if p =1/4 then the average would be 1/p 1/(1/4) or 4.
I'm asking for a formal (but not a full-blooded rigorous) proof of the "obvious".
I "expect" that an average equation is needed ^~^
It's as I thought. Paradoxically, explaining the obvious can take considerable thought!!! The common sense way of "seeing it" is valid fortunately, but it doesn't make good formal mathematics.
The harder cereal box problem was asked later than this easier one. It was answered quite quickly.
The "expect" (cf average), in my last post, was a clue.
This post has been removed by the author.
Looks like no-one's up for this, so I'll put it out of my misery.
By definition, the average (or expected) value for a discrete random
variable X is E(X) = Σ x*P(X=x). In our case, x is the number of trials
until the event occurs = 1*P(1) + 2*P(2) + 3*P(3) + ...
Let p = P(1) be the probability of success in 1 trial and q = 1-p be the
probability of failure in 1 trial. For an event to first occur on the xth
trial, it must have failed to occur during the previous (x-1) trials.
So the probability that the event occurs on the xth trial is:
P(x) = (q^(x-1))p. Let E(X) = m, then:
m = p + 2qp + 3q²p + 4q³p + ...
multiply by q => mq = qp + 2q²p + 3q³p + ...
Subtract that from the previous => m - mq = p + qp + q²p + q³p + ...
=> m(1-q) = p(1 + q + q² + q³ + ...)
But 1 + q + q² + q³ + ... = 1/(1-q) {|q| < 1}
=> m(1-q) = p/(1-q)
=> mp = p/p = 1
=> m = 1/p as required.
I re-edited this since original post on 28 Jan.
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