0.14 More probability

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.