Chapter PIV, Problem 7RE

(a)

To determine

To find: the minimal chance that one will have twins, if no one took a fertility drug.

Answer to Problem 7RE

10.6%

Explanation of Solution

Given:

The probability of females getting twins is $\frac{1}{10}=0.1$

Calculation:

Suppose X is the number of women who have twins while taking drugs for futility.

Success is about making twins.

Failure does not mean having twins.

The trails are independent.

Hence X is binom (10, 0.1)

Let X be the number of pregnant women who, if they do not take fertility drugs, have twins.

The probability of getting twins is $\frac{1}{90}$

The variable X is a binomial $\left(10,\frac{1}{90}\right)$

The probability of minimum one pregnant woman getting twins is

  $\begin{array}{c}P\left(X\ge 1\right)=1-P\left(X=0\right)\\ =1-C{}_0{\left(\frac{1}{90}\right)}^0{\left(\frac{89}{90}\right)}^{10}\\ =0.1057\end{array}$

If both of them are not taking fertility medication, there is a 10.6 percent risk that at least one has twins.

(b)

To determine

To find: the minimal probability that one would have twins, if both of them were to take clomid.

Answer to Problem 7RE

65%

Explanation of Solution

Given:

  $\begin{array}{c}n=10\\ p=0.1\end{array}$

Calculation:

Probability that from 10 pregnant minimum one would have twin is

  $\begin{array}{c}P\left(X\ge 1\right)=1-P\left(X=0\right)\\ =1-C{}_0{\left(0.1\right)}^0{\left(0.9\right)}^{10}\\ =0.6513\end{array}$

There is 65% possibility that among 10 pregnant women minimum one has twin.

(c)

To determine

To find: the probability that if half were taking clomid, at minimum one would have twins.

Answer to Problem 7RE

Explanation of Solution

Given:

  $p=0.1$

Calculation:

The chance that if about half of them take fertility drugs, at least one has twins is,

  $\begin{array}{c}P\left(X\ge 1\right)=1-P\left(\text{no twin from first five}\right)\times P\left(\text{no twin from next five}\right)\\ =1-\left\{C{}_0{\left(\frac{1}{10}\right)}^0{\left(\frac{9}{10}\right)}^5\right\}\times \left\{C{}_0{\left(\frac{1}{90}\right)}^0{\left(\frac{89}{90}\right)}^4\right\}\\ =1-\left(0.543\right)\left(0.4095\right)\\ =1-0.0222\\ =0.9778\end{array}$

If only half of them take fertility drugs, there is a 97.8 percent risk that at least one has twins.