Geometric Distribution If a procedure meets all the conditions of a binomial distribution except that the number of trials is not fixed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by , where p is the probability of success on any one trial. Subjects are randomly selected for the National Health and Nutrition Examination Survey conducted by the National Center for Health Statistics, Centers for Disease Control and Prevention. The probability that someone is a universal donor (with group O and type Rh negative blood) is 0.06. Find the probability that the first subject to be a universal blood donor is the fifth person selected.

Short Answer

Expert verified

The probability that the first subject to be a universal blood donor is the fifth person selected is equal to 0.047.

Step by step solution

01

Given information

The probability that someone is a universal donor is given to be equal to 0.06.

02

Step 2:Required probability

Success is defined as selecting a person who is a universal blood donor.

Let X denote the number of trials at which success is attained, so X follows a geometric distribution.

The probability of success is defined below:

p = 0.06.

The probability of getting the first success at the xth trial is shown below:

Px=p1-px-1

Thus, the probability of the first subject to be a universal blood donor is the fifth person selected (success is attained at the fifthtrial) is computed below:

Px=p1-px-1P5=0.061-0.065-1=0.047

Therefore, theprobability that the first subject to be a universal blood donor is the fifth person selected, which is equal to 0.047.

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Most popular questions from this chapter

In Exercises 5 and 6, refer to the given values, then identify which of the following is most appropriate:discrete random variable, continuous random variable, ornot a random variable.

a. Grades (A, B, C, D, F) earned in statistics classes

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x

P(x)

0

0.031

1

0.156

2

0.313

3

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0.031

Hypergeometric Distribution If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type (such as lottery numbers you selected), while the remaining B objects are of the other type (such as lottery numbers you didn’t select), and if n objects are sampled without replacement (such as six drawn lottery numbers), then the probability of getting x objects of type A and n - x objects of type B is

\(P\left( x \right) = \frac{{A!}}{{\left( {A - x} \right)!x!}} \times \frac{{B!}}{{\left( {B - n + x} \right)!\left( {n - x} \right)!}} \div \frac{{\left( {A + B} \right)!}}{{\left( {A + B - n} \right)!n!}}\)

In New Jersey’s Pick 6 lottery game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 6, B = 43, n = 6, and x = 2.)

For 100 births, P(exactly 56 girls) = 0.0390 and P(56 or more girls) = 0.136. Is 56 girls in 100 births a significantly high number of girls? Which probability is relevant to answering that question?

The accompanying table lists probabilities for the corresponding numbers of girls in four births. What is the random variable, what are its possible values, and are its values numerical?

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P(x)

0

0.063

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0.250

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