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Worked solution

Differentiate the posterior probability over its percentage domain

In another National Eye Centre survey, p% are boys and the rest are girls. The probability a randomly chosen boy wears spectacles is 0.3, and for a randomly chosen girl it is 0.24.

(i) Find p if a randomly chosen child wears spectacles with probability 0.267.

(ii) For general p, let f(p) be the probability that a randomly chosen child who wears spectacles is a girl. Show that f(p)=4(100−p)/(400+p)f(p)=4(100-p)/(400+p). Prove by differentiation that f is decreasing for 0≤p≤1000\le p\le100, and explain the statement in context.

[2017/MI/II/6]

CJC · Discussion · Question 8

Answer

(i) p=45p=45. (ii) f(p)=4(100−p)/(400+p)f(p)=4(100-p)/(400+p) and f′(p)=−2000/(400+p)2<0f'(p)=-2000/(400+p)^2<0 for 0≤p≤1000\le p\le100, with the stated population-composition interpretation.

(i) Use total probability from the source tree

The boy/girl probabilities are p/100p/100 and 1−p/1001-p/100; their spectacle/no-spectacle branches are 0.3/0.7 and 0.24/0.76. Thus (p/100)(0.3)+(1−p/100)(0.24)=0.267(p/100)(0.3)+(1-p/100)(0.24)=0.267. Simplifying gives 0.0006p=0.0270.0006p=0.027, so p=45p=45.

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(ii) Condition on wearing spectacles

f(p)=0.24(1−p/100)0.3(p/100)+0.24(1−p/100).\begin{aligned}&f(p)\\ & =\frac{0.24(1-p/100)}{0.3(p/100)+0.24(1-p/100)}.\end{aligned}

Preserve the source simplification stages:

f(p)=0.24−0.0024p0.003p+0.24−0.0024p=0.0024(100−p)0.0006(400+p)=4(100−p)400+p.\begin{aligned}&f(p)\\ & =\frac{0.24-0.0024p}{0.003p+0.24-0.0024p}\\ & =\frac{0.0024(100-p)}{0.0006(400+p)}\\ & =\frac{4(100-p)}{400+p}.\end{aligned}

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(ii) Differentiate and establish the correct interval

By the quotient rule,

f′(p)=(400+p)(−4)−(400−4p)(400+p)2=−2000(400+p)2.\begin{aligned}&f'(p)\\ & =\frac{(400+p)(-4)-(400-4p)}{(400+p)^2}\\ & =\frac{-2000}{(400+p)^2}.\end{aligned}

On 0≤p≤1000\le p\le100, the denominator is positive, so f′(p)<0f'(p)<0 and f is decreasing. The source’s last line prints “for all real p”; the rational expression is undefined at p=−400. The requested probability domain is the percentage interval[0,100], and the proof here is confined to that interval.

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(ii) Interpret the change in context

As the percentage of boys in the surveyed population increases, the probability that a randomly chosen spectacle-wearing child is a girl decreases. This changes the composition of spectacle wearers; it does not change either within-group wearing probability 0.3 or 0.24.

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Boy probabilityp/100, girl 1−p/100. Given boy: spectacles 0.3, no spectacles 0.7. Given girl: spectacles 0.24, no spectacles 0.76. All six source branch probabilities retained.
Boy probabilityp/100, girl 1−p/100. Given boy: spectacles 0.3, no spectacles 0.7. Given girl: spectacles 0.24, no spectacles 0.76. All six source branch probabilities retained.
Boy probabilityp/100, girl 1−p/100. Given boy: spectacles 0.3, no spectacles 0.7. Given girl: spectacles 0.24, no spectacles 0.76. All six source branch probabilities retained.

Common mistakes

p is a percentage, not a fraction. Do not extend the source’s monotonicity statement to all real numbers across a pole.

Solution notes

Both source asks, all six tree branches, every fraction simplification stage, quotient-rule calculation and contextual conclusion retained. The source’s all-real derivative-domain claim is explicitly restricted to the requested[0,100]; native original preserved. No printed marks.

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