not sure about this one. I know the generalized sum rule but not how to apply it to 100 0.01 disjuncts
This is just the complement of the previous probability you computed: 1-0.99^100, which is indeed approximately 0.632. Rather than compute this directly, you might observe that (1-1/n)^n converges very quickly to 1/e or approximately 0.368.
This is just the complement of the previous probability you computed: 1-0.99^100, which is indeed approximately 0.632. Rather than compute this directly, you might observe that (1-1/n)^n converges very quickly to 1/e or approximately 0.368.
Yeah, nsheppard pointed that out to me after I wrote the fold. Oh well! I’ll know better next time.