Find the probability that at least one valve is defective when two valves are drawn at random.

combinatoricsprobabilitysolution-verification

Problem Statement:

  • A factory A produces $10\ \%$ defective valves
  • and another factory $B$ produces $\mbox{$20\ \%$}$ defective valves.
  • A bag contains $4$ valves of factory $A$
    and $5$ valves of factory B.
  • If two valves are drawn at random from the
    bag, find the probability that at least one valve is defective.

This question has been asked here once.
Find the probability that atleast one valve is defective.

Now, clearly Probability of drawing atleast one defective valve)=1- Probability that both valves drawn are non-defective valves
Using this, the required probability is
$
=1-\left(\frac{\binom{4}{2}}{\binom{9}{2}}(0.9)^2+\frac{\binom{5}{2}}{\binom{9}{2}}(0.8)^2+\frac{\binom{4}{1}\binom{5}{1}}{\binom{9}{2}}(0.9)(0.8)\right)=\frac{517}{1800},
\label{1}\tag{1}
$
as OP has also shown in the above linked post.

However, this is where the confusion arises:
Alternatively, let's consider the following mutually exclusive events:

  1. Both valves are defective and are from factory $A$
  2. Both valves are defective and are from factory $B$
  3. Both valves are defective (one from $A$ and the other from $B$)
  4. One valve is drawn from $A$ and is defective while the other is drawn from $B$ and is non-defective.
  5. One valve is drawn from $B$ and is defective while the other is drawn from $A$ and is non-defective.

Now the required probability = Sum of all the probabilities of mutually exclusive events listed above.

Let's denote probability for the ith event listed above by $P(i)$, where $i=1,2,3,4,5$
$P(1)=\frac{^4C_2}{^9C_2}(0.1)^2\;\;,P(2)=\frac{^5C_2}{^9C_2}(0.2)^2\;\;,P(3)=\frac{^4C_1 \times ^5C_1}{^9C_2}(0.1)(0.2)\;\;$
$P(4)=\frac{^4C_1\times ^5C_1}{^9C_2}(0.1)(0.8)\;\; ,P(5)=\frac{^4C_1 \times ^5C_1}{^9C_2}(0.9)(0.2)\;\;$
Therefore, the required probability is $$P=\sum_{i=1}^{5}P(i)=1/600+1/90+1/90+2/45+1/10=303/1800
\label{2}\tag{2}$$

I want to know why answers in (\ref{1}) and (\ref{2}) above are different. In fact, in the link above, OP had mentioned that answer given in his book was $303/1800$, whereas the comments and answers to that post mentioned the answer as wrong. But $(2)$ clearly shows that there is nothing wrong with the answer. Please help. Thanks.

Best Answer

The error comes from two cases that you are missing:

  • Both valves are drawn from $A$ and exactly one is defective
  • Both valves are drawn from $B$ and exactly one is defective

The probability of one of those events occurring is $$ \frac{\binom{4}{2}}{\binom{9}{2}} \times 0.1 \times 0.9 \times 2 + \frac{\binom{5}{2}}{\binom{9}{2}} \times 0.2 \times 0.8 \times 2 = \frac{54+160}{1800} = \frac{214}{1800} = \frac{517}{1800} - \frac{303}{1800} \,.$$

OP's answer is correct.

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