(→Part(a)) |
(→Part(a)) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 6: | Line 6: | ||
<math> P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} </math> | <math> P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} </math> | ||
| + | |||
Recall geometric series: | Recall geometric series: | ||
| − | |||
| − | for |x| < 1 | + | <math> \sum_{\imath=0}^{\infty} x^{\imath}= \frac{1}{1-x}</math> for |x| < 1 |
| + | |||
| + | |||
| + | <math> P(B) = p\sum_{\imath=0}^{\infty} (1-p)^{4\imath} = \frac{p}{1-(1-p)} </math> | ||
| + | |||
| + | |||
| + | 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.
