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This chapter covers additional principles of probability. After completing this chapter students should be able to: find the probability of a binomial experiment; find the probabilities using Bayes' Formula; find the expected value or payoff in a game of chance; find the probabilities using tree diagrams.

Chapter overview

In this chapter, you will learn to:

  1. Find the probability of a binomial experiment.
  2. Find probabilities using Bayes' Formula.
  3. Find the expected value or payoff in a game of chance.
  4. Find probabilities using tree diagrams.

Binomial probability

In this section, we will consider types of problems that involve a sequence of trials, where each trial has only two outcomes, a success or a failure. These trials are independent, that is, the outcome of one does not affect the outcome of any other trial. Furthermore, the probability of success, p size 12{p} {} , and the probability of failure, 1 p size 12{ left (1 - p right )} {} , remains the same throughout the experiment. These problems are called binomial probability problems. Since these problems were researched by a Swiss mathematician named Jacques Bernoulli around 1700, they are also referred to as Bernoulli trials .

We give the following definition:

Binomial experiment

A binomial experiment satisfies the following four conditions:
  1. There are only two outcomes, a success or a failure, for each trial.
  2. The same experiment is repeated several times.
  3. The trials are independent; that is, the outcome of a particular trial does not affect the outcome of any other trial.
  4. The probability of success remains the same for every trial.

The probability model that we are about to investigate will give us the tools to solve many real-life problems like the ones given below.

  1. If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?
  2. If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 7 out of 10 free throws in a game?
  3. If a medicine cures 80% of the people who take it, what is the probability that among the ten people who take the medicine, 6 will be cured?
  4. If a microchip manufacturer claims that only 4% of his chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective?
  5. If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product?

We now consider the following example to develop a formula for finding the probability of k size 12{k} {} successes in n Bernoulli trials.

A baseball player has a batting average of . 300 size 12{ "." "300"} {} . If he bats four times in a game, find the probability that he will have

  1. four hits
  2. three hits
  3. two hits
  4. one hit
  5. no hits.

Let us suppose S size 12{S} {} denotes that the player gets a hit, and F size 12{F} {} denotes that he does not get a hit.

This is a binomial experiment because it meets all four conditions. First, there are only two outcomes, S size 12{S} {} or F size 12{F} {} . Clearly the experiment is repeated four times. Lastly, if we assume that the player's skillfulness to get a hit does not change each time he comes to bat, the trials are independent with a probability of . 3 size 12{ "." 3} {} of getting a hit during each trial.

We draw a tree diagram to show all situations.

A Tree diagram showing the all the possible situations for this problem.

Let us first find the probability of getting, for example, two hits. We will have to consider the six possibilities, SSFF size 12{ ital "SSFF"} {} , SFSF size 12{ ital "SFSF"} {} , SFFS size 12{ ital "SFFS"} {} , FSSF size 12{ ital "FSSF"} {} , FSFS size 12{ ital "FSFS"} {} , FFSS size 12{ ital "FFSS"} {} , as shown in the above tree diagram. We list the probabilities of each below.

P SSFF = . 3 . 3 . 7 . 7 = . 3 2 . 7 2 size 12{P left ( ital "SSFF" right )= left ( "." 3 right ) left ( "." 3 right ) left ( "." 7 right ) left ( "." 7 right )= left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {}

P SFSF = . 3 . 7 . 3 . 7 = . 3 2 . 7 2 size 12{P left ( ital "SFSF" right )= left ( "." 3 right ) left ( "." 7 right ) left ( "." 3 right ) left ( "." 7 right )= left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {}

P SFFS = . 3 . 7 . 7 . 3 = . 3 2 . 7 2 size 12{P left ( ital "SFFS" right )= left ( "." 3 right ) left ( "." 7 right ) left ( "." 7 right ) left ( "." 3 right )= left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {}

P FSSF = . 7 . 3 . 3 . 7 = . 3 2 . 7 2 size 12{P left ( ital "FSSF" right )= left ( "." 7 right ) left ( "." 3 right ) left ( "." 3 right ) left ( "." 7 right )= left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {}

P FSFS = . 7 . 3 . 7 . 3 = . 3 2 . 7 2 size 12{P left ( ital "FSFS" right )= left ( "." 7 right ) left ( "." 3 right ) left ( "." 7 right ) left ( "." 3 right )= left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {}

P FFSS = . 7 . 7 . 3 . 3 = . 3 2 . 7 2 size 12{P left ( ital "FFSS" right )= left ( "." 7 right ) left ( "." 7 right ) left ( "." 3 right ) left ( "." 3 right )= left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {}

Since the probability of each of these six outcomes is . 3 2 . 7 2 size 12{ left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {} , the probability of obtaining two successes is 6 . 3 2 . 7 2 size 12{6 left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {} .

The probability of getting one hit can be obtained in the same way. Since each permutation has one S size 12{S} {} and three F size 12{F} {} 's, there are four such outcomes: SFFF size 12{ ital "SFFF"} {} , FSFF size 12{ ital "FSFF"} {} , FFSF size 12{ ital "FFSF"} {} , and FFFS size 12{ ital "FFFS"} {} .

And since the probability of each of the four outcomes is . 3 . 7 3 size 12{ left ( "." 3 right ) left ( "." 7 right ) rSup { size 8{3} } } {} , the probability of getting one hit is 4 . 3 . 7 3 size 12{4 left ( "." 3 right ) left ( "." 7 right ) rSup { size 8{3} } } {} .

The table below lists the probabilities for all cases, and shows a comparison with the binomial expansion of fourth degree. Again, p size 12{p} {} denotes the probability of success, and q = 1 p size 12{q= left (1 - p right )} {} the probability of failure.

Outcome Four Hits Three hits Two Hits One hits No Hits
Probability . 3 4 size 12{ left ( "." 3 right ) rSup { size 8{4} } } {} 4 . 3 3 . 7 size 12{4 left ( "." 3 right ) rSup { size 8{3} } left ( "." 7 right )} {} 6 . 3 2 . 7 2 size 12{6 left ( "." 3 right ) rSup { size 8{2} } left ( "." 7 right ) rSup { size 8{2} } } {} 4 . 3 . 7 3 size 12{4 left ( "." 3 right ) left ( "." 7 right ) rSup { size 8{3} } } {} . 7 4 size 12{ left ( "." 7 right ) rSup { size 8{4} } } {}

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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