题目

Problem

Pieces of wood cladding are produced by a timber merchant. There are three types of fault, $A$ , $B$ and $C$ , that can appear in each piece of wood cladding.

The Venn diagram shows the probabilities of a piece of wood cladding having the various types of fault.

A piece of wood cladding is chosen at random.

(a) Find the probability that the piece of wood cladding has more than one type of fault.

(1)

Fault types $A$ and $C$ occur independently.

(b) Find the probability that the piece of wood cladding has no faults.

(4)

Given that the piece of wood cladding has fault $A$ ,

(2)

Two pieces of the wood cladding are selected at random.

(d) Find the probability that both have exactly 2 types of fault.

(3)

解答

解法一

思路

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“more than one type of fault” 包括所有两两重叠和三重重叠。求 $p$ 时用 $A$ 和 $C$ 独立，即 $P(A\cap C)=P(A)P(C)$ 。

答题过程

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The probability of more than one type of fault is

\begin{align*} 0.05+0.04+0.03+0.01=0.13. \end{align*}

From the diagram,

\begin{align*} P(A)=0.10+0.05+0.04+0.01=0.20. \end{align*}

Also,

\begin{align*} P(C)=p+0.04+0.01+0.03=p+0.08. \end{align*}

Since $A$ and $C$ are independent,

\begin{align*} P(A\cap C)=P(A)P(C). \end{align*}

\begin{align*} 0.04+0.01=0.20(p+0.08). \end{align*}

Thus

\begin{align*} 0.05=0.20p+0.016. \end{align*}

Hence

\begin{align*} p=0.17. \end{align*}

The probability of having at least one fault is

\begin{align*} 0.10+0.05+0.08+0.04+0.01+0.03+0.17=0.48. \end{align*}

Therefore the probability of no faults is

\begin{align*} q=1-0.48=0.52. \end{align*}

Given fault $A$ , the probability that the piece also has fault $B$ but not $C$ is

\begin{align*} \frac{0.05}{0.20}=0.25. \end{align*}

The probability that one piece has exactly 2 types of fault is

\begin{align*} 0.05+0.04+0.03=0.12. \end{align*}

For two independent pieces, the probability that both have exactly 2 types of fault is

\begin{align*} (0.12)^2=0.0144. \end{align*}

IAL 2018 Oct Q4

题目

解答

解法一

思路

答题过程