First you need to compute the probability that any one specific coin has nu=0. Call this probability P. Now the number of coins that have nu=0 is itself a binomial distribution with probability P.
You can use the above observation, or you can use a trick: the probability that at least one coin has nu=0 is related in a simple way to the probability that all coins have
Quote:
Originally Posted by i_need_some_help
On part (a), I tried a few different things. Most recently, 1 - binomcdf(1 to 10 | N, mu) ^ (number of coins), but this doesn't seem to be correct.
In the case where mu is 0.05 and we try with one coin, the probability should be 1 - 0.05**10. I can't figure out how to generalize that to multiple coins.
WON'T YOU HELP??
|