Question
Zac raises funds for a library by running a game where players spin a needle. The final position of the needle results in an outcome where a player wins or loses money. The outcomes, with associated probabilities, are shown in the following diagram.
\((a) \ (i)\) Find the expected value of \( X \).
\((ii)\) Interpret your answer to part (a)(i).
To encourage a person to keep playing this game, Zac increases the winning prize for the second game they play from $5 to $6. For each successive game they play, the winning prize continues to increase by $1.
Emily plays \( k \) games. The \( k \)th game is fair.
\((b) \ (i)\) Find the value of \( k \).
\((ii)\) Explain why Zac expects to raise money from the games Emily plays.
▶️ Answer/Explanation
Detailed Solution
Step 1: Calculate the expected value \( E(X) \)
Using the expected value formula:
\[ E(X) = (5 \times 0.40) + (-8 \times 0.1) + (-5 \times 0.2) + (-10 \times 0.3) \]
\[ E(X) = 2 – 0.8 – 1 – 3 = -2.8 \]
Step 2: Interpretation of \( E(X) \)
Any one of the following interpretations is correct:
- On average, players will lose $2.80 per game.
- Players are expected to lose $2.80 per game in the long run.
- This represents the long-term expected average when playing the game many times.
- The expected value is not zero, so the game is not fair.
Step 3: Finding the value of \( k \) (where the game becomes fair)
For the \( k \)th game, we set \( E(X) = 0 \):
\[ (5 + (k – 1) \times 1) \times 0.40 + (-8 \times 0.1) + (-5 \times 0.2) + (-10 \times 0.3) = 0 \]
\[ (4 + k) \times 0.40 + (-8 \times 0.1) + (-5 \times 0.2) + (-10 \times 0.3) = 0 \]
Solving for \( k \):
\[ \frac{2.80}{0.40} + 1 = 8 \]
So, the game becomes fair at \( k = 8 \) games.
Step 4: Explanation of why Zac expects to raise money
\( E(X) < 0 \) for each of the first 7 games, meaning that before the fair game, players are expected to lose money on average. Since most players do not reach the fair game, Zac raises money overall.
……………………………Markscheme……………………………….
Expected Value: \( E(X) = -2.8 \)
Interpretation: Players lose $2.80 per game on average.
Fair Game Value: \( k = 8 \) games.
Fundraising Insight: Zac earns money as \( E(X) < 0 \) for the first 7 games.
Question
A teacher surveys their students to find out if they have eaten at the local Thai and Indian cafés. The results of the survey are shown in the following Venn diagram.
(a) Write down the number of students surveyed.
(b) Write down the number of students who have not eaten at the Indian café.
A student is chosen at random from those surveyed.
(c) Find the probability this student has eaten at both the Thai café and the Indian café.
Let $T$ be the event: a student has eaten at the Thai café.
Let $I$ be the event: a student has eaten at the Indian café.
(d) Find $P(T \cup I)$.
(e) State whether the events $T$ and $I$ are mutually exclusive. Justify your answer.
▶️Answer/Explanation
Detailed solution
(a) Number of Students Surveyed:
The total number of students surveyed can be found by adding up the numbers in all the sections of the Venn diagram.
\[ \text{Total students} = 10 + 13 + 8 + 2 = 33 \]
So, the number of students surveyed is 33.
(b) Number of Students Who Have Not Eaten at the Indian Café:
The students who have not eaten at the Indian café are represented by the area outside the Indian café circle. From the diagram, we can see that 10 students have not eaten at the Indian café.
So, the number of students who have not eaten at the Indian café is 12.
(c) Probability That the Student Has Eaten at Both Cafés:
The students who have eaten at both the Thai and Indian cafés are represented by the intersection of the two circles, which is the number 13 in the Venn diagram.
The probability is given by the ratio of students who have eaten at both cafés to the total number of students surveyed.
\[ P(T \cap I) = \frac{13}{33} \approx 0.394 \text{ or } 39.4\% \]
(d) Probability of $T \cup I$:
The probability of the union of the events $T$ and $I$ is the probability that a student has eaten at the Thai café or the Indian café or both. This is calculated by adding the probabilities of the individual events and subtracting the probability of their intersection.
\[ P(T \cup I) = \frac{10 + 13 + 8 + 2}{33} = \frac{31}{33} \approx 0.939 \text{ or } 93.9\% \]
(e) Are the Events $T$ and $I$ Mutually Exclusive?
Two events are mutually exclusive if they cannot occur at the same time. In this case, the probability of the intersection $P(T \cap I) = 13/33 \neq 0$, meaning the events $T$ and $I$ are not mutually exclusive.
So, no, the events are not mutually exclusive.
……………………………Markscheme……………………………….
(a)
33
(b)
12
(c)
$\frac{13}{33}$ (0.394, 39.4%)
(d)
$P(T \cup I) = \frac{31}{33}$ (0.939, 93.9%)
(e)
$P(T \cap I) \neq 0$ OR $n(T \cap I) \neq 0$
no, they are not mutually exclusive