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 $T$ 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 $q$

It is difficult to diagnose people with type A of the disease and there is an unknown proportion $p$ 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 $\frac{41}{169}$

(b) Find the value of $p$ and the value of $q$ .

(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.

(3)

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

(1)

题目中文翻译

某疾病在人群中分为两种互斥类型。事件 $T$ 表示检测结果为阳性。若患 B 型，阳性概率为 0.96；若未患病，阳性概率为 0.05；若患 A 型，阳性概率为 $q$ 。人群中患 A 型的比例为未知数 $p$ ，患 B 型的比例为 2%。

(a) 补全树状图。

随机选取一人，其检测为阳性的概率是 0.169。对于检测阳性者，未患病的概率为 $\frac{41}{169}$ 。

(b) 求 $p$ 与 $q$ 。

一名医生即将接诊一位已知未患 B 型、但检测结果为阳性的人。

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 $p$ 、0.02、 $0.98-p$ 。第二层每对分支相加为 1。

答题过程

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The first branches are

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

The second branches are

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

for type A,

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

for type B, and

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

for no disease.

(b)

解法一

思路

展开

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

答题过程

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Using $P(T)=0.169$ ,

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

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

Hence

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

Also,

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

Therefore

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

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

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

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

and

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

Substitute into $pq-0.05p=0.1008$ :

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

Thus

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

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

(c)

(i)

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

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

The numerator is

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

The denominator is

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

Therefore

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

So the probability is

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

(ii)

思路

展开

把检测阳性且未患 B 型后的 A 型概率与人群原本的 A 型比例比较。

答题过程

展开

The doctor will find the test useful because the probability of type A is now much higher than the original population probability $0.16$ .

S1 2019 June Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

(i)

思路

答题过程

(ii)

思路

答题过程