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<math> P(B) = p\sum_{\imath=0}^{\infty} (1-p)^{4\imath} = \frac{p}{1-(1-p)} </math> | <math> P(B) = p\sum_{\imath=0}^{\infty} (1-p)^{4\imath} = \frac{p}{1-(1-p)} </math> | ||
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| + | Repeat this for Carol, Ted, and Alice to show that the order of your toss affects your probability of winning. | ||
Latest revision as of 17:09, 9 September 2008
Part(a)
Show that P(B) > P(C) > P(T) > P(A):
- P(H) = p , 0 < p < 1
$ P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} $
Recall geometric series:
$ \sum_{\imath=0}^{\infty} x^{\imath}= \frac{1}{1-x} $ for |x| < 1
$ P(B) = p\sum_{\imath=0}^{\infty} (1-p)^{4\imath} = \frac{p}{1-(1-p)} $
Repeat this for Carol, Ted, and Alice to show that the order of your toss affects your probability of winning.
