Chapter PIV, Problem 9RE

(a)

To determine

To find: the possibility that non would have twins.

Answer to Problem 9RE

0.5905

Explanation of Solution

Given:

  $\begin{array}{c}n=5\\ r=0\end{array}$

Formula used:

  $P\left(X=r\right)=C{}_r{p}^r{\left(1-p\right)}^{n-r}$

Calculation:

Let X be the number of women that she is pregnant and having twins  

A pregnant woman who has taken clomid and has twins has a chance of 0.1

Required Probability is

  $\begin{array}{c}P\left(X=0\right)=C{}_0{\left(0.1\right)}^0{\left(0.9\right)}^5\\ =0.5905\end{array}$

There is a 59 percent probability that pregnant women won't have twins.

(b)

To determine

To find: the probability that exactly one would have twins.

Answer to Problem 9RE

0.3281

Explanation of Solution

Given:

  $\begin{array}{c}n=5\\ r=1\end{array}$

Formula used:

  $P\left(X=r\right)=C{}_r{p}^r{\left(1-p\right)}^{n-r}$

Calculation:

Required Probability is

  $\begin{array}{c}P\left(X=1\right)=C{}_1{\left(0.1\right)}^1{\left(0.9\right)}^4\\ =0.3281\end{array}$

There is a 32.81 percent risk of twins in pregnant women.

(c)

To determine

To find: the probability that minimum three would have twins.

Answer to Problem 9RE

0.8

Explanation of Solution

Given:

  $P\left(X=r\right)=C{}_r{p}^r{\left(1-p\right)}^{n-r}$

Calculation:

Probability for minimum 3 women would have twins is

  $\begin{array}{c}P\left(X\ge 3\right)=P[X=3]+P[X=4]+P[X=5]\\ =C{}_3{\left(0.1\right)}^3{\left(0.9\right)}^2+C{}_4{\left(0.1\right)}^4\left(0.9\right)+C{}_5{\left(0.1\right)}^5{\left(0.9\right)}^0\\ =0.0089\end{array}$

There is only a 0.8 percent probability that twins will be available to at least 3 women.