Q.3.19 - A First Course in Probability

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?

Step-by-Step Solution

Answer

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

a)  $44.29\%$

b) $41.18\%$

Step 1: Given Information (part b)

What percentage of the original class attended the party?

Step 2: 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:

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

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

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

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

=$0.4118$

Step 3: Final Answer (part b)

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

Step 4: Given Information (part a)

what percentage of those attending the party were women? 

Step 5: Explanation (part a)

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

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

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

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

$=0.4429$

Step 6: Final Answer (part a)

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