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 the 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

First calculate(b),then using (b)calculate (a)

a) $44.29\%$

b) $41.18\%$

Step-by-step solution

Given Information (part b)

What percentage of the original class attended the party?

Explanation (part b)

Consider events:

S - a randomly chosen person from the class attends the success party.

M -a randomly chosen person from the class is male

F - a randomly chosen person from the class is female

Given probabilities ( conditioned on being in the class)

$P_c(S\mid F)=0.48$

$P_c(S\mid M)=0.37$

$P_c\left(M\right)=0.62$

Men and women are considered competing hypothesis so:

$F=M^c\Rightarrow P_c\left(F\right)=1-P_c\left(M\right)=38\mathrm{\%}$

$\text{Use the version of Bayes formula (obtained by breaking}S\text{into}SM\text{and}SF\text{)}$

$P_c\left(S\right)=P\left(M\right)P(S\mid M)+P\left(F\right)P(S\mid F)$

$=0.62\cdot 0.37+0.38\cdot 0.48$

=$0.4118$

Final Answer (part b)

The percentage of the original class attended the party is$0.4118$

Given Information (part a)

what percentage of those attending the party were women?

Explanation (part a)

$\text{Therefore, the answer to b) is}41.18\%\text{of people from the class attended the success party.}$

$P_c(F\mid S)=\frac{P_c\left(FS\right)}{P_c\left(S\right)}$

$=\frac{P_c(S\mid F)P_c\left(F\right)}{P_c\left(S\right)}$

$=\frac{0.48\cdot 0.38}{0.4118}$

$=0.4429$

Final Answer (part a)

percentage of those attending the party were women is$0.4429$