Cumulative distribution of binomial random variables

stgermaine

1. The problem statement, all variables and given/known data
The probability of being dealt a full house is approximately 0.0014. Find the probability that in 1000 hands of poker you will be dealt at least 2 full houses


2. Relevant equations
I can use binomial distribution.

3. The attempt at a solution
The probability of getting exactly two full houses would be (1000 choose 2) * (0.0014)^2 * (1 - 0.0014)^ 998

However, because this problem asks for probability of being dealt *at least* two hands, I have to add the probabilities (1000 choose k) * (0.0014)^k * (1 - 0.0014)^ (1000-k), where k is any integers [2, 1000]

Most problems in the textbook (Ross Probability) deal with situations where the possible values of k are only 4-6 integers and they can calculate it by hand, but I think I need to use summation for this problem. How do I do this kind of problems?

awkward

Consider the Poisson distribution.

Ray Vickson

Sometimes when it is too difficult to compute a probability P{A} directly, we instead compute the probability P{B} of the complementary event, then use P{A} = 1 - P{B}.

RGV

stgermaine

Ah I haven't thought of it that way. Thanks!