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.