7 N20/LI/5
Shania and Tina are playing a game. Shania has a bag containing one green disc, red dises and blue
discs, where . A red disc is worth 5 points, a blue disc is worth 2 points and the green disc is worth
0 points. Tina takes two discs from the bag at random. Tina’s score is found by multiplying together
the number of points for each of the two discs she takes.
(a) State Tina’s possible scores.
(b) Show that the expectation of Tina’s score is , and find an expression for the variance
© Given that the variance of Tina’s score is 38 , find the value of .

Ask by Joseph Lawson. in Singapore

Mar 21,2025

Question

7 N20/LI/5
Shania and Tina are playing a game. Shania has a bag containing one green disc, red dises and blue
discs, where . A red disc is worth 5 points, a blue disc is worth 2 points and the green disc is worth
0 points. Tina takes two discs from the bag at random. Tina’s score is found by multiplying together
the number of points for each of the two discs she takes.
(a) State Tina’s possible scores.
(b) Show that the expectation of Tina’s score is , and find an expression for the variance
© Given that the variance of Tina’s score is 38 , find the value of .

Ask by Joseph Lawson. in Singapore

Mar 21,2025

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part (a): State Tina's possible scores.

Tina can draw two discs from the bag, and we need to consider the combinations of discs she can draw. The discs in the bag are:

- 1 green disc (0 points)
- $$ r $$ red discs (5 points each)
- $$ 2r $$ blue discs (2 points each)

The possible combinations of discs and their corresponding scores are:

1. **Two red discs**: 
   - Score = $$ 5 \times 5 = 25 $$

2. **One red disc and one blue disc**: 
   - Score = $$ 5 \times 2 = 10 $$

3. **One red disc and one green disc**: 
   - Score = $$ 5 \times 0 = 0 $$

4. **Two blue discs**: 
   - Score = $$ 2 \times 2 = 4 $$

5. **One blue disc and one green disc**: 
   - Score = $$ 2 \times 0 = 0 $$

6. **Two green discs**: 
   - Not possible since there is only one green disc.

Thus, the possible scores Tina can achieve are $$ 0, 4, 10, $$ and $$ 25 $$.

### Part (b): Show that the expectation of Tina's score is $$ \frac{27 r-11}{3 r+1} $$, and find an expression for the variance.

To find the expectation, we need to calculate the probabilities of each score and then multiply by the score.

1. **Total number of discs**: 
   - Total = $$ 1 + r + 2r = 3r + 1 $$

2. **Calculating probabilities**:
   - **Score 25 (two red discs)**:
     - Probability = $$ \frac{\binom{r}{2}}{\binom{3r+1}{2}} = \frac{\frac{r(r-1)}{2}}{\frac{(3r+1)(3r)}{2}} = \frac{r(r-1)}{(3r+1)(3r)} $$

- **Score 10 (one red and one blue)**:
     - Probability = $$ \frac{r \cdot 2r}{\binom{3r+1}{2}} = \frac{2r^2}{\frac{(3r+1)(3r)}{2}} = \frac{4r^2}{(3r+1)(3r)} $$

- **Score 0 (one red and one green)**:
     - Probability = $$ \frac{r}{\binom{3r+1}{2}} = \frac{r}{\frac{(3r+1)(3r)}{2}} = \frac{2r}{(3r+1)(3r)} $$

- **Score 4 (two blue discs)**:
     - Probability = $$ \frac{\binom{2r}{2}}{\binom{3r+1}{2}} = \frac{r(2r-1)}{\frac{(3r+1)(3r)}{2}} = \frac{2r(2r-1)}{(3r+1)(3r)} $$

Now, we can calculate the expected score $$ E(X) $$:

$$
E(X) = 25 \cdot P(25) + 10 \cdot P(10) + 0 \cdot P(0) + 4 \cdot P(4)
$$

Substituting the probabilities:

$$
E(X) = 25 \cdot \frac{r(r-1)}{(3r+1)(3r)} + 10 \cdot \frac{4r^2}{(3r+1)(3r)} + 0 + 4 \cdot \frac{2r(2r-1)}{(3r+1)(3r)}
$$

Now, let's simplify this expression.

### Part (c): Given that the variance of Tina's score is 38, find the value of $$ r $$.

To find the variance, we need to calculate $$ E(X^2) $$ and use the formula:

$$
Var(X) = E(X^2) - (E(X))^2
$$

We will calculate $$ E(X^2) $$ using the scores and their probabilities, and then set up the equation to find $$ r $$.

Let's perform the calculations for parts (b) and (c) now.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{25\left(r\left(r-1\right)\right)}{\left(\left(3r+1\right)\times 3r\right)}+\frac{10\times 4r^{2}}{\left(\left(3r+1\right)\times 3r\right)}+\frac{4\left(2r\left(2r-1\right)\right)}{\left(\left(3r+1\right)\times 3r\right)}$$
- step1: Remove the parentheses:
  $$\frac{25r\left(r-1\right)}{\left(3r+1\right)\times 3r}+\frac{10\times 4r^{2}}{\left(3r+1\right)\times 3r}+\frac{4\times 2r\left(2r-1\right)}{\left(3r+1\right)\times 3r}$$
- step2: Reduce the fraction:
  $$\frac{25\left(r-1\right)}{3\left(3r+1\right)}+\frac{10\times 4r^{2}}{\left(3r+1\right)\times 3r}+\frac{4\times 2r\left(2r-1\right)}{\left(3r+1\right)\times 3r}$$
- step3: Reduce the fraction:
  $$\frac{25\left(r-1\right)}{3\left(3r+1\right)}+\frac{40r}{3\left(3r+1\right)}+\frac{4\times 2r\left(2r-1\right)}{\left(3r+1\right)\times 3r}$$
- step4: Reduce the fraction:
  $$\frac{25\left(r-1\right)}{3\left(3r+1\right)}+\frac{40r}{3\left(3r+1\right)}+\frac{8\left(2r-1\right)}{3\left(3r+1\right)}$$
- step5: Transform the expression:
  $$\frac{25\left(r-1\right)+40r+8\left(2r-1\right)}{3\left(3r+1\right)}$$
- step6: Multiply the terms:
  $$\frac{25r-25+40r+8\left(2r-1\right)}{3\left(3r+1\right)}$$
- step7: Multiply the terms:
  $$\frac{25r-25+40r+16r-8}{3\left(3r+1\right)}$$
- step8: Calculate:
  $$\frac{81r-33}{3\left(3r+1\right)}$$
- step9: Factor:
  $$\frac{3\left(27r-11\right)}{3\left(3r+1\right)}$$
- step10: Reduce the fraction:
  $$\frac{27r-11}{3r+1}$$
Calculate or simplify the expression $$ 25^2 * (r*(r-1))/((3*r+1)*(3*r)) + 10^2 * (4*r^2)/((3*r+1)*(3*r)) + 4^2 * (2*r*(2*r-1))/((3*r+1)*(3*r)) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{25^{2}\left(r\left(r-1\right)\right)}{\left(\left(3r+1\right)\times 3r\right)}+\frac{10^{2}\times 4r^{2}}{\left(\left(3r+1\right)\times 3r\right)}+\frac{4^{2}\left(2r\left(2r-1\right)\right)}{\left(\left(3r+1\right)\times 3r\right)}$$
- step1: Remove the parentheses:
  $$\frac{25^{2}r\left(r-1\right)}{\left(3r+1\right)\times 3r}+\frac{10^{2}\times 4r^{2}}{\left(3r+1\right)\times 3r}+\frac{4^{2}\times 2r\left(2r-1\right)}{\left(3r+1\right)\times 3r}$$
- step2: Reduce the fraction:
  $$\frac{25^{2}\left(r-1\right)}{3\left(3r+1\right)}+\frac{10^{2}\times 4r^{2}}{\left(3r+1\right)\times 3r}+\frac{4^{2}\times 2r\left(2r-1\right)}{\left(3r+1\right)\times 3r}$$
- step3: Reduce the fraction:
  $$\frac{25^{2}\left(r-1\right)}{3\left(3r+1\right)}+\frac{400r}{3\left(3r+1\right)}+\frac{4^{2}\times 2r\left(2r-1\right)}{\left(3r+1\right)\times 3r}$$
- step4: Reduce the fraction:
  $$\frac{25^{2}\left(r-1\right)}{3\left(3r+1\right)}+\frac{400r}{3\left(3r+1\right)}+\frac{2^{5}\left(2r-1\right)}{3\left(3r+1\right)}$$
- step5: Evaluate the power:
  $$\frac{625\left(r-1\right)}{3\left(3r+1\right)}+\frac{400r}{3\left(3r+1\right)}+\frac{2^{5}\left(2r-1\right)}{3\left(3r+1\right)}$$
- step6: Evaluate the power:
  $$\frac{625\left(r-1\right)}{3\left(3r+1\right)}+\frac{400r}{3\left(3r+1\right)}+\frac{32\left(2r-1\right)}{3\left(3r+1\right)}$$
- step7: Transform the expression:
  $$\frac{625\left(r-1\right)+400r+32\left(2r-1\right)}{3\left(3r+1\right)}$$
- step8: Multiply the terms:
  $$\frac{625r-625+400r+32\left(2r-1\right)}{3\left(3r+1\right)}$$
- step9: Multiply the terms:
  $$\frac{625r-625+400r+64r-32}{3\left(3r+1\right)}$$
- step10: Calculate:
  $$\frac{1089r-657}{3\left(3r+1\right)}$$
- step11: Factor:
  $$\frac{3\left(363r-219\right)}{3\left(3r+1\right)}$$
- step12: Reduce the fraction:
  $$\frac{363r-219}{3r+1}$$
Solve the equation $$ \frac{363r-219}{3r+1} - \left(\frac{27r-11}{3r+1}\right)^2 = 38 $$.
Solve the equation by following steps:
- step0: Solve for $$r$$:
  $$\frac{363r-219}{3r+1}-\left(\frac{27r-11}{3r+1}\right)^{2}=38$$
- step1: Find the domain:
  $$\frac{363r-219}{3r+1}-\left(\frac{27r-11}{3r+1}\right)^{2}=38,r\neq -\frac{1}{3}$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{363r-219}{3r+1}-\left(\frac{27r-11}{3r+1}\right)^{2}\right)\left(3r+1\right)^{2}=38\left(3r+1\right)^{2}$$
- step3: Simplify the equation:
  $$360r^{2}+300r-340=38\left(3r+1\right)^{2}$$
- step4: Move the expression to the left side:
  $$360r^{2}+300r-340-38\left(3r+1\right)^{2}=0$$
- step5: Calculate:
  $$18r^{2}+72r-378=0$$
- step6: Factor the expression:
  $$18\left(r-3\right)\left(r+7\right)=0$$
- step7: Divide the terms:
  $$\left(r-3\right)\left(r+7\right)=0$$
- step8: Separate into possible cases:
  $$\begin{align}&r-3=0\&r+7=0\end{align}$$
- step9: Solve the equation:
  $$\begin{align}&r=3\&r=-7\end{align}$$
- step10: Check if the solution is in the defined range:
  $$\begin{align}&r=3\&r=-7\end{align},r\neq -\frac{1}{3}$$
- step11: Find the intersection:
  $$\begin{align}&r=3\&r=-7\end{align}$$
- step12: Rewrite:
  $$r_{1}=-7,r_{2}=3$$
### Part (a): Tina's Possible Scores
The possible scores Tina can achieve are $$ 0, 4, 10, $$ and $$ 25 $$.

### Part (b): Expectation and Variance
We have shown that the expectation of Tina's score is:

$$
E(X) = \frac{27r - 11}{3r + 1}
$$

For $$ E(X^2) $$, we found:

$$
E(X^2) = \frac{363r - 219}{3r + 1}
$$

The variance $$ Var(X) $$ is given by:

$$
Var(X) = E(X^2) - (E(X))^2
$$

Substituting the expressions we found:

$$
Var(X) = \frac{363r - 219}{3r + 1} - \left(\frac{27r - 11}{3r + 1}\right)^2
$$

### Part (c): Finding the Value of $$ r $$
Given that the variance of Tina's score is 38, we set up the equation:

$$
\frac{363r - 219}{3r + 1} - \left(\frac{27r - 11}{3r + 1}\right)^2 = 38
$$

Solving this equation yields two potential values for $$ r $$:

1. $$ r_1 = -7 $$ (not valid since $$ r > 1 $$)
2. $$ r_2 = 3 $$

Thus, the valid value of $$ r $$ is:

$$
\boxed{3}
$$
Tina's possible scores are 0, 4, 10, and 25. The expectation of her score is $$ \frac{27r - 11}{3r + 1} $$. The variance is calculated using the formula $$ Var(X) = E(X^2) - (E(X))^2 $$. Given that the variance is 38, the value of $$ r $$ is 3.

Upstudy AI Solution

Answer

Solution

The Deep Dive

(a) Tina’s Possible Scores

(b) Expectation and Variance

© Finding Given Variance

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