# 0.14 More probability

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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, $p$ , and the probability of failure, $\left(1-p\right)$ , 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 $k$ successes in n Bernoulli trials.

A baseball player has a batting average of $\text{.}\text{300}$ . 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 $S$ denotes that the player gets a hit, and $F$ 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, $S$ or $F$ . 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 $\text{.}3$ of getting a hit during each trial.

We draw a tree diagram to show all situations.

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

$P\left(\text{SSFF}\right)=\left(\text{.}3\right)\left(\text{.}3\right)\left(\text{.}7\right)\left(\text{.}7\right)={\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$

$P\left(\text{SFSF}\right)=\left(\text{.}3\right)\left(\text{.}7\right)\left(\text{.}3\right)\left(\text{.}7\right)={\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$

$P\left(\text{SFFS}\right)=\left(\text{.}3\right)\left(\text{.}7\right)\left(\text{.}7\right)\left(\text{.}3\right)={\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$

$P\left(\text{FSSF}\right)=\left(\text{.}7\right)\left(\text{.}3\right)\left(\text{.}3\right)\left(\text{.}7\right)={\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$

$P\left(\text{FSFS}\right)=\left(\text{.}7\right)\left(\text{.}3\right)\left(\text{.}7\right)\left(\text{.}3\right)={\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$

$P\left(\text{FFSS}\right)=\left(\text{.}7\right)\left(\text{.}7\right)\left(\text{.}3\right)\left(\text{.}3\right)={\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$

Since the probability of each of these six outcomes is ${\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$ , the probability of obtaining two successes is $6{\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$ .

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

And since the probability of each of the four outcomes is $\left(\text{.}3\right){\left(\text{.}7\right)}^{3}$ , the probability of getting one hit is $4\left(\text{.}3\right){\left(\text{.}7\right)}^{3}$ .

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

 Outcome Four Hits Three hits Two Hits One hits No Hits Probability ${\left(\text{.}3\right)}^{4}$ $4{\left(\text{.}3\right)}^{3}\left(\text{.}7\right)$ $6{\left(\text{.}3\right)}^{2}{\left(\text{.}7\right)}^{2}$ $4\left(\text{.}3\right){\left(\text{.}7\right)}^{3}$ ${\left(\text{.}7\right)}^{4}$

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