问题及解答

Each of 12 refrigerators of a certain type has been returned to a distributor because of the presence of a high-pitched oscillating noise(高音调振荡噪声/高音調振盪雜訊/高音振動ノイズ) when the refrigerator is running. Suppose that 5 of these 12 have defective compressors and the other 7 have less serious problems. If they are examined in random order, let $X$=the number among the first 6 examined that have a defective compressor. Compute the following:

• (a) $P(X=1)$
• (b) $P(X\geqslant 4)$
• (c) $P(1\leqslant X\leqslant 3)$
   

The variable $X$ obeys hypergeometric distribution.

1. The population or set to be sampled consists of $N=12$ individuals.
2. Each individual can be characterized as a success ($S$, here means having less serious problem) or a failure ($F$, here means having a defective compressor), and there are $M=12-5=7$ successes in the population.
3. A sample of $n=6$ individuals is selected without replacement in such a way that each subset of size $n$ is equally likely to be chosen.

Hence

\[
P(X=x)=h(x;n,M,N)=\dfrac{\binom{M}{x}\cdot\binom{N-M}{n-x}}{\binom{N}{n}}=\dfrac{\binom{7}{x}\cdot\binom{12-7}{6-x}}{\binom{12}{6}}
\]

Here, $x$ satisfies $\max\{0,n-N+M\}\leqslant x\leqslant\min\{n,M\}$. That is $\max\{0,6-12+7\}\leqslant x\leqslant\min\{6,7\}$. It is $1\leqslant x\leqslant 6$.

(a)

\[
P(X=1)=h(1;6,7,12)=\dfrac{\binom{7}{1}\cdot\binom{12-7}{6-1}}{\binom{12}{6}}=\frac{7\times 1}{924}\approx 0.00757576
\]

(b)

\[
\begin{split}
P(X\geqslant 4)&=P(X=4)+P(X=5)+P(X=6)=h(4;6,7,12)+h(5;6,7,12)+h(6;6,7,12)\\
&=\dfrac{\binom{7}{4}\cdot\binom{12-7}{6-4}}{\binom{12}{6}}+\dfrac{\binom{7}{5}\cdot\binom{12-7}{6-5}}{\binom{12}{6}}+\dfrac{\binom{7}{6}\cdot\binom{12-7}{6-6}}{\binom{12}{6}}\\
&=\frac{35\times 10}{924}+\frac{21\times 5}{924}+\frac{7\times 1}{924}\\
&=\frac{462}{924}\\
&=0.5
\end{split}
\]

(c)

\[
P(1\leqslant X\leqslant 3)=1-P(X\geqslant 4)\approx 1-0.5=0.5
\]

Also, you can compute it as

\[
\begin{split}
P(1\leqslant X\leqslant 3)&=P(X=1)+P(X=2)+P(X=3)\\
&=h(1;6,7,12)+h(2;6,7,12)+h(3;6,7,12)\\
&=\dfrac{\binom{7}{1}\cdot\binom{12-7}{6-1}}{\binom{12}{6}}+\dfrac{\binom{7}{2}\cdot\binom{12-7}{6-2}}{\binom{12}{6}}+\dfrac{\binom{7}{3}\cdot\binom{12-7}{6-4}}{\binom{12}{6}}\\
&=\dfrac{7\times 1}{924}+\dfrac{21\times 5}{924}+\dfrac{35\times 10}{924}\\
&=\dfrac{462}{924}\\
&=0.5
\end{split}
\]