Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Explanation of Solution

Given:

  $\begin{array}{c}\text{Total number of pathogenic isolates extracted from patients }\left(n\right)=7\\ \text{Percentage of pathogenic isolates that are resistant to antibiotic vancomycin }\left(p\right)\text{ = 30\%}\\ \text{=0}\text{.30 }\end{array}$

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

  $P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}$

The addition principle in binomial distribution X is shown as:

  $P(X\ge x)=P(X=x)+P(X=x+1)+\mathrm{...}+P(X=n)$

The assumptions of the binomial distribution are:

• There are only two outcomes for each trial, which are treated as a success and a failure.
• There are a fixed number of trials and each trial is independent of the other.
• The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

Answer to Problem 1PP

The probability of success $\left(p\right)=0.3$

The total number of trails $\left(n\right)=7$

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin $=30\%$ and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin $p=0.30$ .

The total number of pathogenic isolates extracted from patients $n=7$ .

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

Answer to Problem 1PP

The probability is 0.025.

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

  $\begin{array}{c}P\left(X=5\right)=\left(\begin{array}{c}7\\ 5\end{array}\right){0.3}^5{\left(1-0.3\right)}^{7-5}\\ =0.025\end{array}$

The required probability is 0.025.

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

  $\begin{array}{l}P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}\\ wherex=6,n=7,p=0.3\\ P(X=6)=\left(\begin{array}{c}7\\ 6\end{array}\right){0.3}^6{\left(1-0.3\right)}^{7-6}\\ P(X=6)=7\times {0.3}^6{0.7}^{7-6}=0.0036\end{array}$

Now,

  $\begin{array}{l}P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}\\ wherex=7,n=7,p=0.3\\ P(X=7)=\left(\begin{array}{c}7\\ 7\end{array}\right){0.3}^7{\left(1-0.3\right)}^{7-7}\\ P(X=7)=1\times {0.3}^7{0.7}^{7-7}=0.0002\end{array}$

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

Answer to Problem 1PP

The probability is 0.0288.

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

  $P(X\ge 5)=P(X=5)+P(X=6)+P(X=7)$

It is known that,

  $\begin{array}{l}P(X=5)=0.025\\ P(X=6)=0.0036\\ P(X=7)=0.0002\end{array}$

Now,

  $\begin{array}{c}P(X\ge 5)=P(X=5)+P(X=6)+P(X=7)\\ =0.025+0.0036+0.0002\\ =0.0288\end{array}$

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.