October 2021 Paper 2 Q14

OCR ACurrent spec11 marksDRVs Probability

14 The probability distribution of a random variable \(X\) is modelled as follows.

\[\mathrm{P}(X = x) = \begin{cases} \dfrac{k}{x} & x = 1, 2, 3, 4, \\ 0 & \text{otherwise,} \end{cases}\]

where \(k\) is a constant.

(a) Show that \(k = \dfrac{12}{25}\). [2]

(b) Show in a table the values of \(X\) and their probabilities. [1]

(c) The values of three independent observations of \(X\) are denoted by \(X_1\), \(X_2\) and \(X_3\).
Find \(\mathrm{P}(X_1 \gt X_2 + X_3)\). [3]

In a game, a player notes the values of successive independent observations of \(X\) and keeps a running total. The aim of the game is to reach a total of exactly 7.

(d) Determine the probability that a total of exactly 7 is first reached on the 5th observation. [5]

Mark scheme (a)

Scheme	Marks
\(k\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = 1\)	M1
\(k \times \frac{25}{12} = 1\) or eg \(\frac{25}{3}k = 4\) or \(25k = 12\).
hence \(k = \frac{12}{25}\) AG	A1
	[2]

Notes

M1: Correct equation involving multiple of \(k\)

A1: Must see previous line and answer

Alternative

Scheme	Marks
or verification: \(\frac{12}{25} + \frac{6}{25} + \frac{4}{25} + \frac{3}{25} = 1\)	M1 A1

Mark scheme (c)

Scheme	Marks
\((3, 1, 1)\ \ (4, 1, 1)\ \ (4, 2, 1)\ \ (4, 1, 2)\)	M1
\(\frac{4}{25} \times \left(\frac{12}{25}\right)^2 + \frac{3}{25} \times \left(\frac{12}{25}\right)^2 + \frac{3}{25} \times \frac{6}{25} \times \frac{12}{25} + \frac{3}{25} \times \frac{12}{25} \times \frac{6}{25}\) oe	M1
\(= \dfrac{288}{3125}\) or 0.09216	A1
	[3]

Notes

M1: At least three of these seen or implied. No extras or repeats.

M1: At least two correct terms, no incorrect coefficients; ft their table.
Allow in terms of \(k\)

A1: Allow 0.0922 (3 sf)

Mark scheme (d)

Scheme	Marks
\((1, 1, 1, 1, 3)\) \((1, 1, 1, 2, 2)\)	B1 B1
\(\left(\frac{12}{25}\right)^4 \times \frac{4}{25} \times 5 + \left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2 \times {}^5\mathrm{C}_2\) oe	M1 A1
\(= \dfrac{41472}{390625}\) or 0.10616832	A1
	[5]

Notes

B1B1 for both sets in any order, without extras. Both soi.
B1 for both sets in any order, with extras.

M1: \(\left(\frac{12}{25}\right)^4 \times \frac{4}{25}\) or \(\left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2\) oe seen. Ignore coeffs. ft their table

A1: For either \(\left(\frac{12}{25}\right)^4 \times \frac{4}{25} \times 5\) or \(\left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2 \times {}^5\mathrm{C}_2\) oe ft their table

A1: Allow 0.106 (3 sf)

Alternative method for M1A1

Scheme	Marks
or \(\left(\frac{12}{25}\right)^4 \times \frac{4}{25} \times (4 + 1) + \left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2 \times ({}^4\mathrm{C}_2 + 4)\)	M1 A1

M1: oe

A1: For either \(\left(\frac{12}{25}\right)^4 \times \frac{4}{25} \times (4 + 1)\) or \(\left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2 \times ({}^4\mathrm{C}_2 + 4)\) oe