Chapter 13.6, Problem 38E

(a)

To determine

To find: Theprobability of all $20$ African carrying the disease.

Answer to Problem 38E

The probability of all carrying the disease is $1\times {10}^{-20}$ .

Explanation of Solution

Given:

  $P\left(\text{carrying the disease}\right)=10\%=\frac{1}{10}.$

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

  $\begin{array}{l}P\left(\text{carrying the disease}\right)=10\%=\frac{1}{10}.\\ P\left(\text{not carrying the disease}\right)=1-\frac{1}{10}=\frac{9}{10}.\end{array}$

The probability of all carrying the disease is:

  $\begin{array}{l}P\left(\text{all carry the disease}\right)=C{}_{20}\times {\left(\frac{1}{10}\right)}^{20}\times {\left(\frac{9}{10}\right)}^0.\\ \Rightarrow 1\times {10}^{-20}.\end{array}$

Hence, the probability of all carrying the disease is $1\times {10}^{-20}$ .

(b)

To determine

To find: Theprobability of exactly half have the disease sickle cells Anaemia.

Answer to Problem 38E

The probability of exactly half have the disease is $1.847\times {10}^{-5}$ .

Explanation of Solution

Given: $P\left(\text{carrying the disease}\right)=10\%=\frac{1}{10}.$

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

According to the given:

  $\begin{array}{l}P\left(\text{carrying the disease}\right)=10\%=\frac{1}{10}.\\ P\left(\text{not carrying the disease}\right)=1-\frac{1}{10}=\frac{9}{10}.\end{array}$

The probability of exactly half have the disease is:

  $\begin{array}{l}P\left(\text{exactly half have the disease}\right)=P\left(10\text{ african has disease}\right).\\ \Rightarrow C{}_{10}\times {\left(\frac{1}{10}\right)}^{10}\times {\left(\frac{9}{10}\right)}^0.\\ \Rightarrow 184756\times {\left(\frac{1}{10}\right)}^{10}=1.847\times {10}^{-5}.\end{array}$

Hence, the probability of exactly half have the disease is $1.847\times {10}^{-5}$ .