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That's what I thought but the key says 49/5832
John Mallinckrodt wrote:
On Fri, 20 Apr 2001, J. Manor wrote:
A student asked me this today and I need some help. Suppose you roll
two dice until you get a sum of 11. The probability you will get a sum
of 11 for the first time on the third roll is? Can you help us?
The probability of getting 11 for the first time on the third roll
= (the probability of NOT getting 11 on the first roll)
x (the probability of NOT getting 11 on the second roll)
x (the probability of GETTING 11 on the third roll)
= (the probability of NOT getting 11 on any given roll)^2
x (the probability of GETTING 11 on the any given roll)
= (34/36)^2 x (2/36) = 17^2/18^3 = ~5.0%
It's interesting to note that this is not very different from the
probability of getting an 11 on the first roll (~5.6%) since the
probability of getting an 11 on either the first or the second
roll is so small.
John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm