Chapter 10, Problem 10.41E

(a)

To determine

To find:

The probability that both parents are carriers in each of the conditions for the given data.

Answer to Problem 10.41E

  $p\left[x=3\right]=0.01563$

Explanation of Solution

Given:

Carrier frequency $=1\text{in}27$

The carrier rate $=1\text{in}250$

Concept used:

Formula

The binomial distribution

  $p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

Calculation:

The probability that offspring develop the disease is approximately $=0.25$

That is $p=0.25$

  $\begin{array}{l}q=1-p\\ q=1-0.25\\ q=0.75\end{array}$

Suppose a husband and wife are both carries of the disease and the wife is pregnant on three different occasions

  $n=3$

The binomial distribution

  $\begin{array}{l}p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}\\ p\left[x=k\right]=\left(\begin{array}{l}3\\ k\end{array}\right){\left(0.25\right)}^k{\left(0.75\right)}^{3-k}\end{array}$

The probability that all three children’s is develop Tay - Sachs disease

  $\begin{array}{l}p\left[x=3\right]=\left(\begin{array}{l}3\\ 3\end{array}\right){\left(0.25\right)}^3{\left(0.75\right)}^{3-3}\\ p\left[x=3\right]=\frac{3!}{3!\left(3-3\right)!}{\left(0.25\right)}^3{\left(0.75\right)}^0\\ p\left[x=3\right]=0.01563\end{array}$

(b)

To determine

To find:

The probability that only one child is develop Tay - Sachs disease for the given data.

Answer to Problem 10.41E

  $p\left[x=1\right]=0.02636$

Explanation of Solution

Given:

Carrier frequency $=1\text{in}27$

The carrier rate $=1\text{in}250$

Concept used:

Formula

The binomial distribution

  $p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

Calculation:

The probability that offspring develop the disease is approximately $=0.25$

That is $p=0.25$

  $\begin{array}{l}q=1-p\\ q=1-0.25\\ q=0.75\end{array}$

Suppose a husband and wife are both carries of the disease and the wife is pregnant on three different occasions

  $n=3$

The binomial distribution

  $\begin{array}{l}p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}\\ p\left[x=k\right]=\left(\begin{array}{l}3\\ k\end{array}\right){\left(0.25\right)}^k{\left(0.75\right)}^{3-k}\end{array}$

The probability that only one child is develop Tay - Sachs disease

  $\begin{array}{l}p\left[x=1\right]=\left(\begin{array}{l}3\\ 1\end{array}\right){\left(0.25\right)}^3{\left(0.75\right)}^{3-1}\\ p\left[x=1\right]=\frac{3!}{1!\left(3-1\right)!}{\left(0.25\right)}^3{\left(0.75\right)}^2\\ p\left[x=1\right]=0.02636\end{array}$

(c)

To determine

To find:

The probability that a child is neither parent is of European Jewish descent for the given data.

Answer to Problem 10.41E

  $p\left[x=0\right]=0.006591$

Explanation of Solution

Given:

Carrier frequency $=1\text{in}27$

The carrier rate $=1\text{in}250$

Concept used:

Formula

The binomial distribution

  $p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

Calculation:

The probability that offspring develop the disease is approximately $=0.25$

That is $p=0.25$

  $\begin{array}{l}q=1-p\\ q=1-0.25\\ q=0.75\end{array}$

Suppose a husband and wife are both carries of the disease and the wife is pregnant on three different occasions

  $n=3$

The binomial distribution

  $\begin{array}{l}p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}\\ p\left[x=k\right]=\left(\begin{array}{l}3\\ k\end{array}\right){\left(0.25\right)}^k{\left(0.75\right)}^{3-k}\end{array}$

The probability that a child is neither parent is of European Jewish descent

  $\begin{array}{l}p\left[x=0\right]=\left(\begin{array}{l}3\\ 0\end{array}\right){\left(0.25\right)}^3{\left(0.75\right)}^{3-0}\\ p\left[x=0\right]=\frac{3!}{1!\left(3-0\right)!}{\left(0.25\right)}^3{\left(0.75\right)}^3\\ p\left[x=0\right]=0.006591\end{array}$