AS June 2025 Q2
2. The discrete random variable \(X\) represents the score when a spinner is spun.
The probability distribution of \(X\) is given by
| \(x\) | 2 | 5 | 9 |
|---|---|---|---|
| \(\mathrm{P}(X = x)\) | 0.6 | 0.3 | 0.1 |
Show your working clearly. (4)
A game is played by spinning the spinner twice.
If the two scores are the same, the number of points earned is 0
If the two scores are different, the number of points earned is the sum of the two scores.
Mehmet plays the game 150 times.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(X) = 2 \times 0.6 + 5 \times 0.3 + 9 \times 0.1\ [= 3.6]\) | M1 | 1.1b |
| \(\mathrm{E}(X^2) = 2^2 \times 0.6 + 5^2 \times 0.3 + 9^2 \times 0.1\ [= 18]\) | M1 | 1.1b |
| \(\mathrm{Var}(X) = 18 - 3.6^2\) | M1 | 1.1b |
| \(= \underline{\mathbf{5.04}}\) | A1 | 1.1b |
| (4) |
Notes
M1: Attempt at \(\mathrm{E}(X)\) with at least two correct products (must be seen in part (a))
M1: Attempt at \(\mathrm{E}(X^2)\) with at least two correct products
M1: Use of “\(\mathrm{E}(X^2)\)” – “\([\mathrm{E}(X)]^2\)” with their values
A1: 5.04 oe Working must be shown
Correct answer on its own with no working scores 0 marks.
SC: \(18 - 3.6^2 = 5.04\) on its own scores M0M0M1A1
| Scheme | Marks | AO |
|---|---|---|
| [\(\mathrm{P}(T = 14) =\)] \(2 \times 0.3 \times 0.1 = 0.06\)* | B1cso* | 1.1b |
| (1) |
Notes
B1*: Correct calculation oe and given answer (allow \(0.3 \times 0.1 + 0.1 \times 0.3 = 0.06\)*)
| Scheme | Marks | AO |
|---|---|---|
| Possible point totals: [0,] 7, 11, [14] | M1 | 3.1b |
| [\(\mathrm{P}(T = 0) = 0.6^2 + 0.3^2 + 0.1^2\ [= 0.46]\)] \(\mathrm{P}(T = 7) = 2 \times 0.6 \times 0.3\ [= 0.36]\) \(\mathrm{P}(T = 11) = 2 \times 0.6 \times 0.1\ [= 0.12]\) [\(\mathrm{P}(T = 14) = 2 \times 0.3 \times 0.1 = 0.06\)] | M1 | 1.1b |
| [\(\mathrm{E}(T) = 0 \times 0.46\)] \(+\, 7 \times 0.36 + 11 \times 0.12 + 14 \times 0.06\) | M1 | 1.1b |
| \(= \underline{\mathbf{4.68}}\) | A1 | 1.1b |
| (4) |
Notes
M1: Realising \(T = 7\), 11 (and 14) are needed If extra incorrect totals are stated, then M0
M1: Attempting \(\mathrm{P}(T = 7)\) and \(\mathrm{P}(T = 11)\) (at least one correct or both with missing ×2)
May be embedded in the calculation for \(\mathrm{E}(T)\), eg \(7 \times 2 \times 0.6 \times 0.3 + 11 \times 2 \times 0.6 \times 0.1 + \ldots\)
M1: Attempting \(\mathrm{E}(T)\) for their values with at least 2 non-zero products for two of the \(T\) values correct or correct ft eg \(7 \times 0.18 + 7 \times 0.18\) only counts as 1 product
Must be for their totals, simply calculating \(\mathrm{E}(X) = 2 \times 0.6 + 5 \times 0.3 + 9 \times 0.1\) here is M0
A1: 4.68 oe Correct answer with no obvious incorrect working scores 4 out of 4
| Scheme | Marks | AO |
|---|---|---|
| \(Y \sim \mathrm{B}(150, 0.06)\) | M1 | 3.3 |
| \(\approx \mathrm{Po}(9)\) | M1 | 1.1b |
| \(\mathrm{P}(Y = 4) \approx 0.0337\) | A1 | 3.4 |
| (3) | ||
| (12 marks) |
Notes
M1: Selecting the correct binomial model (may be implied by sight of \(\mathrm{Po}(9)\))
M1: Writing or using \(\mathrm{Po}(9)\) allow ft \(\mathrm{Po}(np)\) from their stated binomial distribution
A1: awrt 0.0337
SC: awrt 0.0313 from exact binomial scores M1M0A0