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In this chapter, you will learn to:
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:
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.
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
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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