Question

An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it is when she is a nonsmoker. If 32 percent of women of childbearing age are smokers, what percentage of women having ectopic pregnancies are smokers?

Short answer

The 48.48 percentage of women having ectopic pregnancies are smokers.

Step-by-step solution

Given Information

Given that an ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it is when she is a nonsmoker

32 percent of women of childbearing age are smokers

We have to find percentage of women having ectopic pregnancies are smokers

Explanation

The objective is to find the percentage of women having ectopic pregnancies who are smokers.

Let E denote the event that the woman has an ectopic pregnancy.

Let S denote the event that the woman is a smoker.

Let $S^c$denote the event that the woman is a non-smoker.

The known probabilities are,

$P\left(S\right)=32\%$

$\Rightarrow P\left(s^c\right)=1-P\left(s\right)$

$=1-0.32$

=0.68

Consider that an ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it is when she is a smoker.

That is,

$P(E\mid S)=2\times P\left(E\mid S^e\right)$

Probability of Women

The probability of women having ectopic pregnancies are smokers is,

$P(S\mid E)=\frac{P(E\cap S)}{P\left(E\right)}$

$=\frac{P(E\mid S)\times P\left(S\right)}{P\left(E\right)}\left(\begin{array}{c}\text{By the formula of}\\ \text{conditional probability}\end{array}\right)$

$=\frac{P(E\mid S)\times P\left(S\right)}{P(E\mid S)\times P\left(S\right)+P\left(E\mid S^c\right)\times P\left(S^c\right)}$$\left(\begin{array}{c}\text{From the multiplication}\\ \text{rule of probability,}\\ P\left(E\right)=\left[\begin{array}{l}P(E\mid S)\times P\left(S\right)\\ +P\left(E\mid S^c\right)\times P\left(S^c\right)\end{array}\right]\end{array}\right]$

$=\frac{2\times P\left(E\mid S^c\right)\times P\left(S\right)}{2\times P\left(E\mid S^c\right)\times P\left(S\right)+P\left(E\mid S^C\right)\times P\left(S^c\right)}$$\left(\text{Since}P\right(E\mid S)=2*P(E\mid S\left)\right)$

$=\frac{P\left(E\mid S^c\right)[2\times P(S\left)\right]}{P\left(E\mid S^c\right)\left[2\times P\left(S\right)+P\left(S^c\right)\right]}$

$=\frac{2\times P\left(s\right)}{2=P\left(s\right)+P\left(s^2\right)}$

$=\frac{2\times 0.32}{2\times 0.32+0.68}$

$=\frac{0.64}{0.64+0.68}$

=0.4848

$=48.48\%$

Thus, there is a $48.48\%$of chance that the women having ectopic pregnancies are smokers.

Final Answer

There is a 0.4848 of chance that the women having ectopic pregnancies are smokers.