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IAL 2019 June Q3

A Level / Edexcel / S1

IAL 2019 June Paper · Question 3

题目

Problem

A certain disease occurs in a population in 2 mutually exclusive types.

A test has been developed to help diagnose whether or not a person has the disease. The event TT represents a positive result on the test. After a large-scale trial of the test, the following information was obtained.

For a person with type B of the disease the probability of a positive test result is 0.96

For a person who does not have the disease the probability of a positive test result is 0.05

For a person with type A of the disease the probability of a positive test result is qq

It is difficult to diagnose people with type A of the disease and there is an unknown proportion pp of the population with type A.

It is easier to diagnose people with type B of the disease and it is known that 2% of the population have type B.

(a) Complete the tree diagram.

(2)

The probability of a randomly selected person having a positive test result is 0.169

For a person with a positive test result, the probability that they do not have the disease is 41169\frac{41}{169}

(b) Find the value of pp and the value of qq.

(7)

A doctor is about to see a person who she knows does not have type B of the disease but does have a positive test result.

(c) (i) Find the probability that this person has type A of the disease.

(3)

(ii) State, giving a reason, whether or not the doctor will find the test useful.

(1)

解答

(a)

解法一

思路

展开

第一层是 type A、type B、no disease，所以概率分别是 pp、0.02、0.98−p0.98-p。第二层每对分支相加为 1。

答题过程

展开

The first branches are

p,0.02,0.98−p.\begin{align*} p,\quad 0.02,\quad 0.98-p. \end{align*}

The second branches are

q,1−q,\begin{align*} q,\quad 1-q, \end{align*}

for type A,

0.96,0.04,\begin{align*} 0.96,\quad0.04, \end{align*}

for type B, and

0.05,0.95,\begin{align*} 0.05,\quad0.95, \end{align*}

for no disease.

(b)

解法一

思路

展开

先用 P(T)=0.169P(T)=0.169 建立一个含 p,qp,q 的方程。再用“positive 且 no disease”的条件概率求 pp，最后代回求 qq。

答题过程

展开

Using P(T)=0.169P(T)=0.169,

pq+0.02(0.96)+(0.98−p)(0.05)=0.169.\begin{align*} pq+0.02(0.96)+(0.98-p)(0.05)=0.169. \end{align*}

So

pq+0.0192+0.049−0.05p=0.169.\begin{align*} pq+0.0192+0.049-0.05p=0.169. \end{align*}

Hence

pq−0.05p=0.1008.\begin{align*} pq-0.05p=0.1008. \end{align*}

Also,

P(no disease∣T)=41169.\begin{align*} P(\text{no disease}\mid T)=\frac{41}{169}. \end{align*}

Therefore

(0.98−p)(0.05)0.169=41169.\begin{align*} \frac{(0.98-p)(0.05)}{0.169}=\frac{41}{169}. \end{align*}

Since 0.169=16910000.169=\frac{169}{1000},

(0.98−p)(0.05)=0.041.\begin{align*} (0.98-p)(0.05)=0.041. \end{align*}

So

0.98−p=0.82,\begin{align*} 0.98-p=0.82, \end{align*}

and

p=0.16.\begin{align*} p=0.16. \end{align*}

Substitute into pq−0.05p=0.1008pq-0.05p=0.1008:

0.16q−0.05(0.16)=0.1008.\begin{align*} 0.16q-0.05(0.16)=0.1008. \end{align*}

Thus

0.16q=0.1088,\begin{align*} 0.16q=0.1088, \end{align*}

so

q=0.68.\begin{align*} q=0.68. \end{align*}

(c)

解法一

思路

展开

已知不是 type B 且 positive，所以分母只包括 type A positive 与 no disease positive。分子是 type A positive。

答题过程

展开

We need

P(type A∣T and not type B).\begin{align*} P(\text{type A}\mid T\text{ and not type B}). \end{align*}

The numerator is

pq=0.16(0.68)=0.1088.\begin{align*} pq=0.16(0.68)=0.1088. \end{align*}

The denominator is

pq+(0.98−p)(0.05)=0.1088+0.041.\begin{align*} pq+(0.98-p)(0.05)=0.1088+0.041. \end{align*}

Therefore

0.10880.1088+0.041=0.7263….\begin{align*} \frac{0.1088}{0.1088+0.041}=0.7263\ldots. \end{align*}

So the probability is

0.726.\begin{align*} 0.726. \end{align*}

The doctor will find the test useful because the probability of type A is now much higher than the original population probability 0.160.16.