Cumulative distribution of binomial random variables

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stgermaine
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Homework Statement


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


Homework Equations


I can use binomial distribution.

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?
 
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Consider the Poisson distribution.
 
stgermaine said:

Homework Statement


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


Homework Equations


I can use binomial distribution.

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?

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