The problem is based on a Venn diagram representing a group of 50 students. The Venn diagram shows the number of students who wear glasses (G), trainers (T), and have a mobile phone (M). The problem has four parts: (i) Describe, using set notation, the region that contains only one student. (ii) Find $n(T' \cap (G \cup M))$. (iii) Find the probability that a randomly selected student wears trainers but does not wear glasses. (iv) Find the probability that two students picked at random from those wearing trainers both have mobile phones.

1. Problem Description

The problem is based on a Venn diagram representing a group of 50 students. The Venn diagram shows the number of students who wear glasses (G), trainers (T), and have a mobile phone (M). The problem has four parts:

(i) Describe, using set notation, the region that contains only one student.

(ii) Find $n(T' \cap (G \cup M))$.

(iii) Find the probability that a randomly selected student wears trainers but does not wear glasses.

(iv) Find the probability that two students picked at random from those wearing trainers both have mobile phones.

2. Solution Steps

(i) The region with one student is outside of the three circles (G, T, M), which represents the students who do not wear glasses, trainers or have a mobile phone. This region can be expressed as the complement of the union of G, T, and M:

$(G \cup T \cup M)'$

(ii) We want to find $n(T' \cap (G \cup M))$. This represents the number of students who are not wearing trainers (T') and either wear glasses or have a mobile phone (or both), i.e., $G \cup M$.

$T' \cap (G \cup M) = (G \cup M) - (T \cap (G \cup M))$

$n(T' \cap (G \cup M)) = n(G \cup M) - n(T \cap (G \cup M))$.

$G \cup M$ contains all students who wear glasses or have a mobile phone or both. $n(G \cup M) = 0+2+19+2+14+9 = 46$.

$T' \cap (G \cup M)$ can also be written as $(T' \cap G) \cup (T' \cap M) = (G - (G \cap T)) \cup (M - (M \cap T))$.

From the Venn Diagram, $T' \cap G$ is made up of the section labeled 0 and 19 so $n(T' \cap G) = 0 + 19 = 19$. Also from the Venn Diagram, $T' \cap M$ is made up of 9, and 19 so $n(T' \cap M) = 9 + 19 = 28$. However, we must remove those students that belong to $T' \cap G \cap M = 19$, so the number will be $19 + 28 - 19 = 28$.

However, it seems easier to count the number of individuals who satisfy the condition:

$T'$ are those students outside the T circle. These numbers are $0, 19, 9, 1, G \cap M, G only, M only, \text{outside}$. The set is made up of students with the following attributes: glasses and mobile phone only:2, glass only: 0, mobile phone only: 9, neither: 1, glass only and mobile phone:

1

9. $G \cup M$ means the students in G or M or both. The regions are 0, 2, 19, 2, 14, 9

The Intersection means those that satisfy both of these. This includes $0, 2, 19, 2, 9$. Add them together: $0 + 2 + 19 + 9 = 30$. Also included the value of 1 who is outside both T, G, and M. Thus we have $30+1 = 31 - 3 = 28$.

Another way is as follows:

$T'$ means not in $T$. That means, take the whole set and remove the numbers that are in the region $T$ or $3+2+14+1 = 20$. Thus $n(T') = 50-20 = 30$.

$G \cup M$ means $0 + 2 + 19 + 2 + 14 + 9 = 46$.

To find $T' \cap (G \cup M)$, we can calculate $n(T' \cap (G \cup M)) = n(T') + n(G \cup M) - n(T' \cup (G \cup M)) = 30 + 46 - n(T' \cup (G \cup M))$.

Let's find it by counting directly: Students not in T: 0, 19, 9,

1. We are looking for the number of students who are in $T'$ AND in $G \cup M$. In $T'$ AND $G \cup M$.

Therefore, we want those in $G \cup M$ but not in $T$. $G \cup M$ contains 0, 2, 19, 2, 14,

9. We do not want the overlap in T. $3, 2, 14, 2$

Students NOT in T: $0, 19, 9, 1$. Students in G or M: $0, 2, 2, 19, 14, 9$

Students NOT in T AND (in G or M): $0 + 19 + 9 = 28$. Also consider outside the circles, which is 1, but neither have glasses or mobile phone or use trainers, so this is not included.

(iii) The probability that a student wears trainers but does not wear glasses is $P(T \cap G')$.

From the Venn diagram, the number of students who wear trainers is $3 + 2 + 14 + 1 = 20$. The number of students wearing trainers and glasses is $2 + 2 = 4$. The number of students that only wear trainers and don't use glasses is those wearing only trainers: $3+14=17$. Those outside of glasses and inside T is $3+14=17$. So the number of students in T, not in G is $n(T) - n(T \cap G) = (3+2+14+1) - (2+2) = 20-4 = 16$. Then the probability that this student wears trainers but does not wear glasses is $\frac{17}{50}$.

The solution given is 21/50, but the number of those wearing only trainers $3+14 = 17$. If you include the students that wear trainers but do not have glasses i.e. have mobile phones then we add 14 so $3+14 = 17$. Thus the probability is $\frac{17}{50}$.

(iv) The total number of students wearing trainers is $3+2+14+1 = 20$. The number of students wearing trainers who have mobile phones is $14 + 2 = 16$.

We want to find the probability that two students picked at random from those wearing trainers both have mobile phones.

The probability that the first student has a mobile phone is $\frac{16}{20} = \frac{4}{5}$.

Given that the first student has a mobile phone, there are now 19 students wearing trainers, of whom 15 have mobile phones.

The probability that the second student has a mobile phone is $\frac{15}{19}$.

Therefore, the probability that both students have mobile phones is $\frac{16}{20} \times \frac{15}{19} = \frac{4}{5} \times \frac{15}{19} = \frac{60}{95} = \frac{12}{19}$.

3. Final Answer

(i) $(G \cup T \cup M)'$

(ii) $28$

(iii) $\frac{17}{50}$

(iv) $\frac{12}{19}$