问题及解答

A company that produces fine crystal knows from experience that $10\%$ of its goblets have cosmetic flaws and must be classified as "seconds."

• (a) Among six randomly selected goblets, how likely is it that only one is a second?
• (b) Among six randomly selected goblets, what is the probability that at least two are seconds?
• (c) If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?
   

Let $N$ and $S$ denote the two states of selected goblet. $N$ means "normal", $S$ means "second". 

Let $X$ be the number of "second" goblets among six randomly selected ones.

(a) There are six possible outcomes for $X=1$.

\[
\begin{aligned}
NNNNNS\\
NNNNSN\\
NNNSNN\\
NNSNNN\\
NSNNNN\\
SNNNNN
\end{aligned}
\]

Hence, the probability is

\[
P(X=1)=b(1;6,0.1)=\binom{6}{1}\cdot 0.1^1\cdot(1-0.1)^{6-1}=0.354294
\]

(b)

\[
\begin{split}
P(2\leqslant X)&=1-P(X < 2)=1-P(X\leqslant 1)\\
&=1-\sum_{y=0}^{1}b(y;6,0.1)\\
&=1-\bigl[b(0;6,0.1)+b(1;6,0.1)\bigr]\\
&=1-\biggl[\binom{6}{0}0.1^0\cdot(1-0.1)^{6-0}+\binom{6}{1}0.1^1\cdot(1-0.1)^{6-1}\biggr]\\
&=1-\bigl[0.531441+0.354294\bigr]\\
&=0.114265
\end{split}
\]

Or, compute $P(2\leqslant X)=P(2\leqslant X\leqslant 6)=\sum_{y=2}^{6}b(y;6,0.1)$.

(c)

Since the goblets are selected one by one, the two outcomes $NNNNN$ and $NNNNS$ are combined into $NNNN$. Thus, the other four outcomes are

\[
\begin{aligned}
NNNSN\\
NNSNN\\
NSNNN\\
SNNNN
\end{aligned}
\]

Therefore, the probability is

\[
(1-0.1)^4+\binom{4}{1}\cdot 0.1^1\cdot(1-0.1)^4=0.91854
\]