Standard +0.8 This is a Further Maths statistics question requiring students to set up a hypergeometric probability equation and solve a resulting cubic equation. While the hypergeometric setup is standard, solving for the population parameter m from a given probability requires algebraic manipulation of factorials/binomial coefficients and solving a non-trivial equation, making it moderately challenging but within reach for Further Maths students.
7 A bag contains \(2 m\) yellow and \(m\) green counters. Three counters are chosen at random, without replacement. The probability that exactly two of the three counters are yellow is \(\frac { 28 } { 55 }\).
Determine the value of \(m\).
Use \({}^{2m}C_2\) and \(m\); Divide by \({}^{3m}C_3\)
\[= \frac{2m(2m-1)}{2} \times m \div \frac{3m(3m-1)(3m-2)}{6}\]
\[= \frac{2m(2m-1)}{(3m-1)(3m-2)}\]
A1
Correct expression in terms of \(m\) (allow with \(m\) not cancelled yet)
\[\frac{2m(2m-1)}{(3m-1)(3m-2)} = \frac{28}{55}\]
M1
Equate to \(\frac{28}{55}\) & simplify to three-term quadratic
\[\Rightarrow 16m^2 - 71m + 28 = 0\]
A1
Correct simplified quadratic, or (quadratic) \(\times m = 0\), aef
\(m = 4\) BC; Reject \(m = \frac{7}{16}\) as \(m\) is an integer
M1, A1 [7]
Solve to get both \(4\) and \(\frac{7}{16}\); Explicitly reject \(m = \frac{7}{16}\)
OR: \(\dfrac{2m(2m-1) \times m \times 3!}{3m(3m-1)(3m-2) \times 2}\) then as above
Multiplication method can get full marks, but if no \(3\) or \(3!\), max M1M0A0 M1A0M0A0
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# Question 7:
$$\frac{{}^{2m}C_2 \times m}{{}^{3m}C_3}$$ | **M1, M1** | Use ${}^{2m}C_2$ and $m$; Divide by ${}^{3m}C_3$
$$= \frac{2m(2m-1)}{2} \times m \div \frac{3m(3m-1)(3m-2)}{6}$$ | |
$$= \frac{2m(2m-1)}{(3m-1)(3m-2)}$$ | **A1** | Correct expression in terms of $m$ (allow with $m$ not cancelled yet)
$$\frac{2m(2m-1)}{(3m-1)(3m-2)} = \frac{28}{55}$$ | **M1** | Equate to $\frac{28}{55}$ & simplify to three-term quadratic
$$\Rightarrow 16m^2 - 71m + 28 = 0$$ | **A1** | Correct simplified quadratic, or (quadratic) $\times m = 0$, aef
$m = 4$ **BC**; Reject $m = \frac{7}{16}$ as $m$ is an integer | **M1, A1** [7] | Solve to get both $4$ and $\frac{7}{16}$; Explicitly reject $m = \frac{7}{16}$
**OR:** $\dfrac{2m(2m-1) \times m \times 3!}{3m(3m-1)(3m-2) \times 2}$ then as above | | Multiplication method can get full marks, but if no $3$ or $3!$, max M1M0A0 M1A0M0A0
The image you've shared only shows the **contact/back page** of an OCR document — specifically the OCR address, customer contact details, and a note about call recording. It does not contain any mark scheme questions, answers, or mark allocations.
To extract mark scheme content, please share the **actual mark scheme pages** containing the questions and their corresponding answers/marks. I'd be happy to format those for you once the correct pages are provided.
7 A bag contains $2 m$ yellow and $m$ green counters. Three counters are chosen at random, without replacement. The probability that exactly two of the three counters are yellow is $\frac { 28 } { 55 }$.
Determine the value of $m$.
\hfill \mbox{\textit{OCR Further Statistics AS 2020 Q7 [7]}}