Binomial Expansion: Standard B\((n,p)\) probabilities — NJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
The random variable \(Y\) has distribution \(\mathrm{B}(12, p)\) for \(0 < p < 1\).
- Show that \(\dfrac{\mathrm{P}(Y=k)}{\mathrm{P}(Y=k-1)} = \dfrac{(13-k)\,p}{k(1-p)}\).
It is given that the mode of the distribution is 4.
- Find the exact range of values of \(p\).
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(i) \[\begin{aligned} \frac{\mathrm{P}(Y=k)}{\mathrm{P}(Y=k-1)} &= \frac{\dbinom{12}{k}p^k(1-p)^{12-k}}{\dbinom{12}{k-1}p^{k-1}(1-p)^{12-k+1}}\\[6pt] &= \frac{\dfrac{12!}{k!(12-k)!}\,p}{\dfrac{12!}{(k-1)!(13-k)!}\,(1-p)}\\[6pt] &= \frac{(13-k)\,p}{k(1-p)} \end{aligned}\] (Shown)
(ii)
Mode is 4, so \(\mathrm{P}(Y=4) > \mathrm{P}(Y=3)\) and \(\mathrm{P}(Y=4) > \mathrm{P}(Y=5)\).
From \(\mathrm{P}(Y=4) > \mathrm{P}(Y=3)\): \[\begin{aligned} \frac{9p}{4(1-p)} &> 1\\[4pt] 9p &> 4-4p\\[4pt] p &> \frac{4}{13} \end{aligned}\]
From \(\mathrm{P}(Y=4) > \mathrm{P}(Y=5)\), equivalently \(\dfrac{\mathrm{P}(Y=5)}{\mathrm{P}(Y=4)} < 1\): \[\begin{aligned} \frac{8p}{5(1-p)} &< 1\\[4pt] 8p &< 5-5p\\[4pt] p &< \frac{5}{13} \end{aligned}\]