In a school survey, a group of 80 students are asked about how much time per week (to nearest hour) they spend on their co-curricular activities (CCA). The readings are shown below: (The table is given in figure_description.) A student is selected random from the group. Defining the events as follows: G : The student is a girl. L : The student spends 6 hours or less weekly. M : The student spends 4 hours or more weekly. Find the following probabilities in terms of $k$.

1. Determine the categories representing the event $L' \cup M'$.

The event $L$ is spending $6$ hours or less, so its complement $L'$ is spending more than $6$ hours, which corresponds to the category '7 or more'.
The event $M$ is spending $4$ hours or more, so its complement $M'$ is spending less than $4$ hours, which corresponds to the category '3 or less'.
Thus, $L' \cup M'$ includes all students in the '3 or less' and '7 or more' categories.

2. Find the total number of students in $L' \cup M'$ and calculate the probability.

$\text{Number of students in } L' \cup M' = 17 + 18 + 10 + k = 45 + k$

$\mathrm{P}(L' \cup M') = \frac{45 + k}{80}$

1. Determine the number of outcomes for $L'$ and $G \cap L'$.

Since $L$ is the event that a student spends 6 hours or less, the complementary event $L'$ is that a student spends 7 hours or more.
From the table, the total number of students in $L'$ is:
$n(L') = 10 + k$

The event $G \cap L'$ refers to the students who are girls and spend 7 hours or more. From the table, this is:
$n(G \cap L') = k$

2. Calculate the conditional probability $P(G \mid L')$.

Using the conditional probability ratio based on the number of outcomes:
$P(G \mid L') = \frac{n(G \cap L')}{n(L')} = \frac{k}{10 + k}$

1. Formulate an equation using the given probability of $P(L \cap M)$ to find $k$.

The event $L \cap M$ corresponds to the students who spend exactly 4 to 6 hours weekly.
The total number of such students is $20 + (15 - k) = 35 - k$.
Given $P(L \cap M) = \frac{2}{5}$ and the total number of students is 80:
$\frac{35 - k}{80} = \frac{2}{5}$
$35 - k = 80 \times \frac{2}{5}$
$35 - k = 32$
$k = 3$

2. Calculate the individual probabilities $P(L)$ and $P(M)$ using the found value of $k$.

With $k = 3$, the total number of students in event $L$ (6 hours or less) is the sum of the first two columns:
$17 + 18 + 20 + (15 - 3) = 67$
So, $P(L) = \frac{67}{80}$.

The total number of students in event $M$ (4 hours or more) is the sum of the last two columns:
$20 + (15 - 3) + 10 + 3 = 45$
So, $P(M) = \frac{45}{80} = \frac{9}{16}$.

3. Check the condition for independence, $P(L \cap M) = P(L)P(M)$, to conclude.

For $L$ and $M$ to be independent, $P(L \cap M)$ must equal $P(L) \times P(M)$.
$P(L) \times P(M) = \frac{67}{80} \times \frac{9}{16} = \frac{603}{1280}$
We know $P(L \cap M) = \frac{2}{5} = \frac{512}{1280}$.
Since $P(L \cap M) \neq P(L) \times P(M)$, the events $L$ and $M$ are not independent.

1. Identify the number of students corresponding to the intersection of events $G$ and $(L \cap M)$.

The event $L \cap M$ represents students who spend $\le 6$ hours ($L$) AND $\ge 4$ hours ($M$), which is exactly the '4 to 6' hours category.
The intersection $G \cap (L \cap M)$ represents girls who spend $4$ to $6$ hours weekly.
From the given table, the number of such students is $15 - k$.

2. Use the property of mutually exclusive events to solve for $k$.

If $G$ and $(L \cap M)$ are mutually exclusive, their intersection is empty, meaning the probability is $0$.
$\text{P}(G \cap (L \cap M)) = 0$
$\frac{15 - k}{80} = 0$
$15 - k = 0$
$k = 15$