The random variable X has the negative binomial distribution NB(3, p) .

Let \(f(x)\) denote the probability that X takes the value x .

(i)     Write down an expression for \(f(x)\) , and show that

\[\ln f(x) = 3\ln \left( {\frac{p}{{1 - p}}} \right) + \ln (x - 1) + \ln (x - 2) + x\ln (1 - p) - \ln 2{\text{ .}}\]

(ii)     State the domain of f .

(iii)     The domain of f is extended to \(]2,{\text{ }}\infty [\) . Show that

\(\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x - 1}} + \frac{1}{{x - 2}} + \ln (1 - p){\text{ .}}\)

Jo has a biased coin which has a probability of 0.35 of showing heads when tossed. She tosses this coin successively and the \({3^{{\text{rd}}}}\) head occurs on the \({Y^{{\text{th}}}}\) toss. Use the result in part (a)(iii) to find the most likely value of Y .

(i)     \(f(x) = \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  2
\end{array}} \right){p^3}{(1 - p)^{x - 3}}\)     M1A1

Note: Award M1A0 for \(f(x) = \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  2
\end{array}} \right){p^3}{q^{x - 3}}\)

taking logs,     M1

\(\ln f(x) = \left( {\ln \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  2
\end{array}} \right){p^3}(1 - p){}^{x - 3}} \right)\)

\( = \ln \left( {\frac{{(x - 1)(x - 2)}}{2} \times {p^3}{{(1 - p)}^{x - 3}}} \right)\)     A1

Note: Award A1 for simplifying binomial coefficient, seen anywhere.

\( = \ln \left( {\frac{{(x - 1)(x - 2)}}{2} \times {p^3}\frac{{{{(1 - p)}^x}}}{{{{(1 - p)}^3}}}} \right)\)     A1

Note: Award A1 for correctly splitting \({{{(1 - p)}^{x - 3}}}\) , seen anywhere.

\( = 3\ln \left( {\frac{p}{{1 - p}}} \right) + \ln (x - 1) + \ln (x - 2) + x\ln (1 - p) - \ln 2\)     AG

(ii)     the domain is {3, 4, 5, …}     A1

Note: Do not accept \(x \geqslant 3\)

(iii)     differentiating with respect to x ,     M1

\(\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x - 1}} + \frac{1}{{x - 2}} + \ln (1 - p)\)     AG 

[7 marks]

setting \(f'(x) = 0\) and putting p = 0.35 ,

\(\frac{1}{{x - 1}} + \frac{1}{{x - 2}} + \ln 0.65 = 0\)     M1A1

solving, x = 6.195…     A1

we need to check x = 6 and 7

f (6) = 0.1177… and f (7) = 0.1148…     A1

the most likely value of Y is 6     A1

Note: Award the final A1 for the correct conclusion even if the previous A1 was not awarded.

[5 marks]

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being \(x \geqslant 3\) . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being \(x \geqslant 3\) . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.