Question

A total of 48 percent of the women and 37 percent of the men who took a certain “quit smoking” class remained nonsmokers for at least one year after completing the class. These people then attended a success party at the end of a year. If 62 percent of the original class was male,

(a) what percentage of those attending the party were women?

(b) what percentage of the original class attended the party?

Short answer

$\text{a.The percentageofthoseattendingthepartywerewomenis}44.29\%\text{.}$

$\text{b.The percentageoftheoriginalclassattendedthepartyis}41.18\%\text{.}$

Step-by-step solution

Given information 

Probability of women that attended the party is 0.38.

So, the probability of women that attended the party is 1-0.38=0.62

The probability that the person attended the party given she is woman is 0.48.

The probability that the person attended the party given she is woman is 0.37

Calculation of solution (Part a)

$\text{The percentage of women that attended the party can be calculated as:}$

$P(\text{Women}\mid \text{Attended the party})=\frac{P\left(\text{Women and Attended the party}\right)}{P(\text{Attended the party)}}$

$=\frac{0.48(0.38)}{0.48(0.38)+(0.37)(0.62)}$

$=0.4429$

$=44.29\%$

role="math" localid="1646971453310" $\text{Therefore, the required percentage is}44.29\%\text{.}$

Calculation of solution (Part b)

$\text{The percentage of original class that attended the party can be calculated as:}$

$P\left(\begin{array}{l}\text{Attended the}\\ \text{party}\end{array}\right)=P\left(\begin{array}{l}\text{Attended the}\\ \text{party|Women}\end{array}\right)P\left(\text{Women}\right)+P\left(\begin{array}{c}\text{Attended the}\\ \text{party}|Men\end{array}\right)P\left(Men\right)$

$=0.48(0.38)+0.37(0.62)$

$=0.4118$

$=41.18\%$

$\text{Thus, the required percentage is}41.18\%\text{.}$