Cumulative distribution of binomial random variables

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Homework Help Overview

The problem involves calculating the probability of being dealt at least 2 full houses in 1000 hands of poker, using the binomial distribution with a given probability of 0.0014 for a full house.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the binomial distribution and the need for summation to find the probability of at least 2 full houses. One participant suggests considering the Poisson distribution as an alternative approach.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. One participant expresses gratitude for a new perspective on the problem, indicating that the conversation is productive.

Contextual Notes

Participants note that typical textbook problems involve fewer integers for k, making calculations more straightforward, which raises questions about handling larger sums in this context.

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!
 

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