S1 June 2013 Q5
5. A biased die with six faces is rolled. The discrete random variable \(X\) represents the score on the uppermost face. The probability distribution of \(X\) is shown in the table below.
| \(x\) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(a\) | \(a\) | \(a\) | \(b\) | \(b\) | 0.3 |
A biased die with five faces is rolled. The discrete random variable \(Y\) represents the score which is uppermost. The cumulative distribution function of \(Y\) is shown in the table below.
| \(y\) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| \(\mathrm{F}(y)\) | \(\dfrac{1}{10}\) | \(\dfrac{2}{10}\) | \(3k\) | \(4k\) | \(5k\) |
Each die is rolled once. The scores on the two dice are independent.
| Scheme | Marks |
|---|---|
| \(3a + 2b = 0.7\) | M1 |
| \(a + 2a + 3a + 4b + 5b + 1.8 = 4.2\) or \(6a + 9b = 2.4\) | M1 |
| \(5b = 1\) Attempt to solve | M1 |
| \(b = \underline{\mathbf{0.2}}\) cao | B1 |
| \(a = \underline{\mathbf{0.1}}\) cao | B1 |
| (5) |
Notes
Probabilities outside [0, 1] should be awarded M0
1st M1 for an attempt at a linear equation in \(a\) and \(b\) based on sum of probs. = 1
2nd M1 for an attempt at a second linear equation in \(a\) and \(b\) based on \(\mathrm{E}(X) = 4.2\) Allow one slip.
3rd M1 for an attempt to solve their 2 linear equations based on sum of probs and \(\mathrm{E}(X)\). Must reduce to a linear equation in one variable. 1st B1 for \(b\) and 2nd B1 for \(a\). Answers only score B1B1 only. The 3rd M1 may be implied if M2 is scored and both correct answers are given.
ALT B1B1 for stating \(b\) and \(a\).
1st M1 for showing that sum of probs. = 1
2nd M1 for showing that \(\mathrm{E}(X) = 4.2\)
3rd M1 for an overall comment “(therefore) \(a = \ldots\)and \(b = \ldots\)” No comment loses this mark.
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = 1\times 0.1 + 2^2\times 0.1 + 3^2\times 0.1 + 4^2\times 0.2 + 5^2\times 0.2 + 6^2\times 0.3\ (= 20.4)\) (*) | B1cso |
| (1) |
Notes
B1cso for a fully correct expression (no incorrect work seen). E.g. allow \(14\times 0.1 + 41\times 0.2 + 36\times 0.3\) Or \(0.1 + 0.4 + 0.9 + 3.2 + 5 + 10.8\). Allow in a table (with 20.4) but without “+” explicitly seen.
| Scheme | Marks |
|---|---|
| \([\mathrm{Var}(X) =]\ 20.4 - 4.2^2 \quad [= 2.76]\) | M1 |
| \(\mathrm{Var}(5 - 3X) = 9\,\mathrm{Var}(X)\) | M1 |
| \(= \underline{\mathbf{24.84}}\) or 24.8 (allow \(\frac{621}{25}\)) cao | A1 |
| (3) |
Notes
1st M1 for a correct expression for \(\mathrm{Var}(X)\). Must see \(-4.2^2\)
2nd M1 for \((-3)^2\,\mathrm{Var}(X)\) or better, no need for a value. Accept \(-3^2\) if it clearly is used as +9 later.
| Scheme | Marks |
|---|---|
| \([5k = 1\) so] \(k = \underline{\mathbf{0.2}}\) | B1 |
| (1) |
| Scheme | Marks | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\mathrm{P}(Y = 1) = 0.1\) | B1 | ||||||||||||
| e.g. \(\mathrm{P}(Y = 2) = \mathrm{F}(2) - \mathrm{F}(1) = 0.1\) | M1 | ||||||||||||
Ignore incorrect or no label if table fully correct | A1 | ||||||||||||
| (3) |
Notes
B1 for \(\mathrm{P}(Y = 1) = 0.1\)
M1 for correct use of \(\mathrm{F}(y)\) to find one other prob. Can ft their \(k\) if finding \(\mathrm{P}(Y = y)\) for \(y \gt 2\). Can be implied by one other prob. correct or correct ft. Look out for \(\mathrm{P}(3) = 3k - 0.2\) or \(\mathrm{P}(4) = \mathrm{P}(5) = k\).
A1 for a fully correct probability distribution. Correct table only is 3/3
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X = 1)\times\mathrm{P}(Y = 1)\) ,= 0.01 cao | M1, A1 |
| (2) | |
| (15 marks) |
Notes
M1 for a correct expression or answer ft their \(\mathrm{P}(Y = 1)\) and their \(\mathrm{P}(X = 1)\)
A1 for 0.01 or exact equivalent only
Don’t ISW here e.g. \(0.1\times 0.1 + 0.1\times 0.1\) or \(2\times 0.1\times 0.1\) are M0A0