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From a batch of $100$ items of which $20$ are defective, exactly two items are chosen, one at a time, without replacement. Calculate the probability that the second item chosen is defective.
(a) $\dfrac{2}{5}$
(b) $\dfrac{19}{100}$
(c) $\dfrac{1}{5}$
(d) None of these

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Last updated date: 19th Jun 2024
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Answer
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Hint: The event of choosing the second item as defective is comprised of two sub events. One is when the first chosen item is good, and the other when the first chosen item is defective. Using the multiplication rule of probability, we can determine the probabilities of these two events. And the final probability will be equal to the sum of the probabilities of the two events.

Complete step by step solution:
The total number of items in the batch is equal to $100$ items of which $20$ are defective. This means that the number of good items is equal to $80$.
According to the question, exactly two items are chosen without replacement, and the second item is defective. The term “without replacement” means that the first item is not put back into the batch before choosing the second item. Since there are two kinds of items in the batch, one defective and the other good, there will be two cases of the event of choosing two items without replacement from the batch. Let us consider these two cases separately.
Case I: The first item is good and the second item is defective.
The probability of choosing a good item from the $80$ defective items from the batch which contains a total of $100$ items is equal to $\dfrac{80}{100}$.
Now, since one item has been chosen, so we are left with $99$ items. But since the good item was chosen so the number of defective items is still the same, that is, $20$. So the probability of choosing the second item as defective is equal to $\dfrac{20}{99}$. By the multiplication rule of probability, the probability of the occurrence of this case is given by
$\Rightarrow {{p}_{1}}=\dfrac{80}{100}\times \dfrac{20}{99}........(i)$
Case II: Both items are defective.
The probability of choosing a defective item from the $20$ defective items from the batch which contains a total of $100$ items is equal to $\dfrac{20}{100}$.
Now, same as the previous case, the total number of items in the batch becomes $99$. But since a defective item is already chosen, so now we are left with $19$ defective items. So the probability of choosing the second item as defective is equal to $\dfrac{19}{99}$. Therefore, by the multiplication theorem the probability of occurrence of this case is given by
$\Rightarrow {{p}_{2}}=\dfrac{20}{100}\times \dfrac{19}{99}........(ii)$
Therefore, the total probability of choosing the second item as defective is given by
$\begin{align}
  & \Rightarrow p={{p}_{1}}+{{p}_{2}} \\
 & \Rightarrow p=\dfrac{80}{100}\times \dfrac{20}{99}+\dfrac{20}{100}\times \dfrac{19}{99} \\
 & \Rightarrow p=\dfrac{20}{100\times 99}\left( 80+19 \right) \\
 & \Rightarrow p=\dfrac{20}{100\times 99}\left( 99 \right) \\
 & \Rightarrow p=\dfrac{20}{100} \\
 & \Rightarrow p=\dfrac{1}{5} \\
\end{align}$
Hence, the correct answer is option (c).

Note: Do not think that the choosing second item as defective means that the first item must be good. The first choice is completely independent so any event can happen. Also, since in this case items are chosen without replacement, the total number of items will not be fixed.