C HAPTER 8 Section 8.2 – The Geometric Distribution.

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C HAPTER 8 Section 8.2 – The Geometric Distribution

T HE G EOMETRIC D ISTRIBUTION Recall, that in the case of a binomial random variable, the number of trials is fixed beforehand, and the binomial variable X counts the number of successes in a fixed number of trials. For example, the probability of getting 5 heads in 6 tosses would be represented as binomPdf(6,.5, 5) By way of comparison, there are situations in which the goal is to obtain a fixed number of successes. In particular, if the goal is to obtain one success, a random variable X can be defined that counts the number of trials needed to obtain that first success. A random variable that satisfies this description is called geometric. The distribution produced by this random variable is called a geometric distribution. The possible values of a geometric random variable are an infinite set because it is theoretically possible to proceed indefinitely without ever having a success. Note that a geometric distribution starts at P(X=1), not at P(X=0) like a binomial distribution. This is because starting at 0 would be implying that there could be 0 tries in obtaining the first success.

EXAMPLES OF GEOMETRIC SITUATIONS Flip a coin until you get heads. Roll a die until you get a 3. In basketball, attempt a three-point shot until you make a basket.

T HE G EOMETRIC S ETTING Definition: A geometric setting arises when we perform independent trials of the same chance process and record the number of trials until a particular outcome occurs. The four conditions for a geometric setting are B inary? The possible outcomes of each trial can be classified as “success” or “failure.” I ndependent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial. T rials? The goal is to count the number of trials until the first success occurs. S uccess? On each trial, the probability p of success must be the same. B I T S

E XAMPLE 8.15 – R OLL A DIE Suppose you roll a die repeatedly until you get a 3. Is this a geometric setting? The event of interest is rolling a 3 so this would be binary (success/failure) Each toss is independent of the others The random variable is defined as X = the number of trials until a 3 occurs The probability of success is the same for each toss This is a geometric setting since all the conditions were met See example 8.15 on p.465 for further explanation

E XAMPLE 8.16 – D RAW AN A CE Suppose you repeatedly draw cards without replacement from a deck of 52 cards until you draw an ace. Is this a geometric setting? The event of interest is drawing an ace so this is binary (success/failure) The draws would not be independent because you do not replace the previous draw This is not a geometric setting since the draws are not independent. See example 8.16 on p.465 for further explanation.

R ULES FOR C ALCULATING G EOMETRIC P ROBABILITIES Using example 8.16, P(X = 1) = P(success on 1 st roll) = 1/6 P(X = 2) = P(success on 2 nd roll) = (5/6)(1/6) P(X = 3) = P(success on 3 rd roll) = (5/6)(5/6)(1/6) In general, this can be modeled as: If X has the geometric distribution with probability p of success and (1 – p ) of failure on each trial, the possible values of X are 1, 2, 3, …. If n is any one of these values, Geometric Probability

G EOMETRIC D ISTRIBUTION G RAPH The following is a graph of the geometric distribution that represents the probability of a success after X of rolls. Geometric distributions will always be skewed to the right since each time a trial goes by without a success you are multiplying by a number less than 1 (the probability of a failure)

T HE E XPECTED V ALUE AND O THER P ROPERTIES OF THE G EOMETRIC R ANDOM V ARIABLE If you flip a fair coin, how many times would you expect to have to flip it to observe the first head? 2 times If you are rolling a die, how many times would you expect to have to roll it to observe your first 3? 6 times If X is a geometric random variable with probability p of success on each trial, then its mean (expected value) is E ( X ) = µ X = 1/ p. The variance of X is (1 – p )/ p 2 If X is a geometric random variable with probability p of success on each trial, then its mean (expected value) is E ( X ) = µ X = 1/ p. The variance of X is (1 – p )/ p 2 The Mean and Standard Deviation of Geometric Random Variable

E XAMPLE 8.18

“M ORE ” T HAN A C ERTAIN N UMBER OF T RIALS

E XAMPLE 8.19 – A PPLYING THE F ORMULA

Homework: P.468-476 #’s 37, 38, 42, 44 & 49