Re: Statistics Question

[Bernard Cleyet <georgeann@REDSHIFT.COM>]

Multiply individual probabilities when one is using "and"; add them with
"or" -- e.g.  prob. of any single number with a throw of one die is 1/6;
to get the same number again is again 1/6,  To get both is 1/36;  to get a
particular number in one of two throws is 1/3.

since 11 is obtained by either 5 -- 6  or 6 -- 5   the first rolls is 2 *
1/36)  the prob. of not (desired) is 1- (2/36)  second roll is (and) so
(1-2/36)^2;   third roll is "and ", also, but with 2/36 so, if I haven't
goofed, is ((1-2/36)^2)*(2/36).


bc

I'm no mathematician, so though this is an easy probability problem, I
likely have goofed -- I await correction.

"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?