Q3.15 - A First Course in Probability

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? 

Step-by-Step Solution

Answer

The 48.48 percentage of women having ectopic pregnancies are smokers.

Step 1 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  

Step 2 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,

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

Step 3: 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.

Step 4 Final Answer

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