Chapter 5

Chapter 5--Probability Distributions

Section 5-1 

Random variable- A variable whose values are determined by chance. (has a single numerical value).

Probability distribution-The values a random variable can assume and the corresponding probabilities of the values.

Discrete random variable- A variable that takes values from an finite or countable set.

Continuous random variable- A variable that takes values from a infinite or uncountable set. 

Example: Identify as discrete random variable, continuous random variable, or not a random variable.

• 1. Exact weights of quarters-continuous since there is an infinite number of different weights. ie. is there another weight between 2.6 grams and 2.7 grams? yes, and between any other two weights given.

  2. Number of tosses of a quarters required to get a head-discrete since you can list all the possible outcomes, 1 or 2 or 3 or...

  3. Responses to the survey question “Did you smoke at least one cigarette in the last week?” not a random variable since it does not have a numerical value, just a yes or no answer.

  4. Exact foot lengths-continuous since there is an infinite number of outcomes.

Requirements of a probability distribution:

1)    The sum of the probabilities is 1

2)    Each probability is between 0 and 1.

5-1 #28 Expected Value in Roulette.  When playing roulette at the Venetian casino in Las Vegas, a gambler is trying to decide whether to bet $5 on the number 27 or $5 that the outcome is any one of these 5 possibilities: 0, 00, 1, 2, 3.  From example 6, we know that the expected value of the $5 bet for a single number is $ -.26. For the $5 bet that the outcome is 0, 00, 1, 2, or 3, there is a probability of 5/38 of making a net profit of $30 and a 33/38 probability of losing $5.

Expected Value Video for further explanation 

Section 5-2

Binomial Probability Distribution Requirements

• 1. Fixed number of trials

  2. Trials are independent

  3. Two categories (Success/Failure)

  4. Constant probabilities 

Binomial Formula:

P(x) = nCx  * p^x   * q^(n-x)

p= probability of success

q= probability of failure (1-p)

n= number of trials

x= number of successes among n trials

Example:  Determine whether the given procedure results in a binomial distribution.

• 1. In an Idaho potato commission survey of 1000 adults, subjects are asked to select their favorite vegetables, and each response was recorded as “potatoes” or “other”.

Binomial, as all 4 of the above requirements are satisfied.

2.  Ten different senators are randomly selected without replacement, and the number of terms that they have served are recorded.

Not Binomial, there are more than 2 outcomes (number of terms they have served)

5-2 #13 Guessing Answers.  Standard tests, such as the SAT, ACT, or Medical College Admission Test (MCAT), typically use multiple choice questions, each with five possible answers (a,b,c,d,e), one of which is correct. Assume you guess the answers to the first 3 questions.

              A. Use the multiplication rule to find the probability that the first two guesses are wrong and the third is correct.  That is, find P(WWC), where W denotes a wrong answer and C denotes a correct answer.

P(WWC)=P(W)*P(W)*P(C)=(4/5)*(4/5)*(1/5)=16/125

B.  Beginning with WWC, make a complete list of the different possible arrangements of two wrong answers and one correct answer, then find the probability of each entry in the list.

P(WWC)=P(W)*P(W)*P(C)=(4/5)*(4/5)*(1/5)=16/125

P(WCW)=P(W)*P(C)*P(W)=(4/5)*(1/5)*(4/5)=16/125

P(CWW)=P(C)*P(W)*P(W)=(1/5)*(4/5)*(4/5)=16/125

              C. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made?

P(exactly one correct) = P(1)=P(WWC) + P(WCW) + P(CWW) = 16/125 + 16/125 + 16/125 = 3(16/125) = .384 

Binomial Formula:

P(x) = nCx  * p^x   * q^(n-x)

p= probability of success

q= probability of failure (1-p)

n= number of trials

x= number of successes among n trials

This is a shortcut to the above example (#13).  Redoing part C using this formula would be

P(x) = nCx  * p^x   * q^(n-x)

P(1) = 3C1  *(1/5)^1  *  (4/5)^2

=3*(1/5)(4/5)(4/5) Notice we have the same pattern as in #13 ...3 combinations of 16/125 

=3(16/125)=.384

Assume that random guesses are made for eight multiple choice questions on an SAT test, so that there are n=8 trials, each with probability of success (correct) given by p=0.20. Find the probability for the number of correct answers. Note: You can use your calculator or statcrunch for these.

SUPER HELPFUL video on the binomial distribution and how to use Statcrunch to solve them

5-2 #28  Based on a Harris poll, among adults who regret getting tattoos, 20% say they were too young when they got their tattoos.  Assume that five adults who regret getting tattoos are randomly selected, and find the indicated probability.