The random variable X has the distribution \({\text{B}}(n{\text{ , }}p)\) .

(a)     (i)     Show that \(\frac{{{\text{P}}(X = x)}}{{{\text{P}}(X = x - 1)}} = \frac{{(n - x + 1)p}}{{x(1 - p)}}\) .

(ii)     Deduce that if \({\text{P}}(X = x) > {\text{P}}(X = x - 1)\) then \(x < (n + 1)p\) .

(iii)     Hence, determine the value of x which maximizes \({\text{P}}(X = x)\) when \((n + 1)p\) is not an integer.

(b)     Given that n = 19 , find the set of values of p for which X has a unique mode of 13.

(a)     (i)     \(\frac{{{\text{P}}(X = x)}}{{{\text{P}}(X = x - 1)}} = \frac{{\left( {\frac{{n!}}{{(n - x)!x!}} \times {p^x} \times {{(1 - p)}^{n - x}}} \right)}}{{\left( {\frac{{n!}}{{(n - x + 1)!(x - 1)!}} \times {p^{x - 1}} \times {{(1 - p)}^{n - x + 1}}} \right)}}\)     M1A1

\( = \frac{{(n - x + 1)p}}{{x(1 - p)}}\)     AG

(ii)     if \({\text{P}}(X = x) > {\text{P}}(X = x - 1)\) then

\((n - x + 1)p > x(1 - p)\)     (M1)A1

\(np - xp + p > x - px\)     A1

\(x < (n + 1)p\)     AG

(iii)     to maximise the probability we also need

\({\text{P}}(X = x) > {\text{P}}(X = x + 1)\)     (M1)

\(\frac{{\left( {n - (x + 1) + 1} \right)p}}{{(x + 1)(1 - p)}} < 1\)

\(np - xp < x - xp + 1 - p\)

\(p(n + 1) < x + 1\)     A1

hence \(p(n + 1) > x > p(n + 1) - 1\)     (A1)

so x is the integer part of \((n + 1)p\) i.e. the largest integer less than \((n + 1)p\)     A1

[9 marks]

(b)     the mode is the value which maximises the probability     (R1)

\(20p > 13 > 20p - 1\)     M1

\( \Rightarrow p > \frac{{13}}{{20}} = 0.65\), and \(p < \frac{7}{{10}} = 0.70\)     A1A1

(it follows that \(0.65 < p < 0.7\))

[4 marks]

Total [13 marks]

Many candidates made a reasonable attempt at (a)(i) and (ii) but few were able to show that the mode is the integer part of \((n + 1)p\). Part (b) also proved difficult for most candidates with few correct solutions seen.