Solution verification with conditional probability problem

Question

I want to check my solutions for this problem:

After a weekend seminar, some participants go to a restaurant together on the last evening. $53\%$ of the male seminar participants take part, for the female seminar participants the corresponding part is $38\%$. What is the probability that a woman will enter the restaurant first, assuming that the restaurant-goers appear in random order, one at a time, and that $66\%$ of all seminar participants are male?

Give a suitable model with probability space and calculate with conditional probabilities

So for the model $(\Omega, F,P)$ I have written

$\Omega=\{M,M^c\} \times \{R,R^c\}$ where

$M=$ male seminar participants, $M^c=$ female seminar participants,$R=$ restaurant-goer,$R=$ non restaurant-goer

$F=2^{\Omega}$ and

$P: F \rightarrow [0,1], (m,r) \rightarrow [0,1]$ where $m \in \{M,M^c\}$,$r \in \{R,R^c\}$

So I have for $P(m,r)=P(r|m)P(m)$

From the problem we have $P(R|M)=0,53 \\P(R|M^c)=0,38 \\ P(M)=0,66 \\ P(M^c)=0,33$

We want to find $P(M^c|R)=\frac{P(R|M^c)P(M^c)}{P(R|M)*P(M)+P(R|M^c)*P(M^c)}=0,27$

Is everything correct or have I done any mistake? I appreciate any help :)

Answer 1

need to fix this: $P(M^c) = 0.34$ because $100-66=34$.

It seems to me $\Omega = \{M,M^c\} \times \{R,R^c\}$ is not defined correctly.

Suppose there are 10 participants, 6 are male, 4 are female. Let's represent the 6 male participants by $m_1, m_2, m_3, m_4, m_5, m_6$, and the 4 female participants by $f_1, f_2, f_3, f_4$.

Suppose 3 male participants and 2 female participants went to the restaurant. Let's represent the attribute (i.e. characteristic of the participant) of going to the restaurant by $r$ and going home by $h$.

$\Omega = \{ (m_1, r), (m_2, r), (m_3, r)$, $(m_4, h ), (m_5, h)$, $(m_6, h), (f_1,r), (f_2, r)$ $(f_3, h), (f_4, h)\}$.

Notice: if $A$ is the set of male participants and $B$ is the set of female participants then

$( A \cup B ) \times \{ r, h \} = \{ (m_1,r), (m_2,r), (m_3,r), (m_4,r), (m_5,r), (m_6,r), (f_1,r), (f_2, 4), (f_3, r), (f_4, r), (m_1,h), (m_2,h), (m_3,h), (m_4,h), (m_5,h), (m_6,h), (f_1,h), (f_2, h), (f_3, h), (f_4, h) \}$.

Do you see how this set includes many extra outcomes, such as $(m_1, h), (m_4,r), (f_1,h), (f_3,r)$?

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I made the same mistake. I wanted to show a different way of defining an outcome $\omega = (i, s, d )$ where $i \in \{ 1,2, \dots, n \}, s \in \{ m,f \}, d \in \{ r,h \}$. $\:$ (there are $n$ participants)

Then I got fancy and decided to use the cross product. $\Omega = X \times S \times D$. But I realized if $(1,m,r)$ is a possible outcome then $(1,m,h)$ is not a possible outcome. If Bob went to the restaurant then Bob did not go home.

It is okay to specify an outcome as an ordered pair $(m_1, r)$ but in this problem you need to be careful how to specify the set of all outcomes $\Omega$.

Answer 2

Let $M$ and $F$ be the total number of males and females, respectively, on the seminar. Let $N$ be the total number of participants in the seminar. Let $M_R$ and $F_R$ be the number of male and females, respectively who go to restaurant.

Here we have then that $\frac{M_R}{M}=0.53$, $\frac{F_R}{F}=0.38$, $\frac{M}{N}=\frac23$ (just used 2/3 instead of 0.66) and $\frac{F}{N}=\frac13$ (just used 1/3 instead of 0.33). Hence we have \begin{align} M_R&=0.53\times\frac23\times N\\ F_R&=0.38\times\frac13\times N\\ \end{align} Therefore the proobability that a female enters the restaurant first is $$\frac{F_R}{F_R+M_R}=\frac{0.38\times\frac13\times N}{0.38\times\frac13\times N + 0.53\times\frac23\times N}=\frac{19}{72}\approx0.2639.$$

Note. I used the fractions 1/3 and 2/3 so that we have the sum being 1.