Question

Let A consists of 3 good and 2 defective articles. Let B consists of 4 good and 1 defective articles. A new lot C is formed by taking 3 articles from A and 2 from B. Find the probability that an article chosen at random from C is defective.
A. \[\dfrac{1}{3}\]
B. \[\dfrac{2}{5}\]
C. \[\dfrac{8}{{25}}\]
D. None

Answer 1

Hint: We will first find the individual probabilities of A and B and the values of of $P\left( {\dfrac{D}{A}} \right)$ and $P\left( {\dfrac{D}{B}} \right)$ then put all these in the formula for the probability we require and thus have the answer.

Complete step-by-step answer:
We are given that: A consists of 3 good and 2 defective items and B consists of 4 good and 1 defective item.
$P\left( {\dfrac{D}{A}} \right)$ means the probability of getting a defective item while picking random item from slot A and similarly $P\left( {\dfrac{D}{B}} \right)$ means the probability of getting a defective item while picking random item from slot B.
Therefore, $P\left( {\dfrac{D}{A}} \right)$ will be equal to the number of defective items in A divided by the total number of items in A.
So, $P\left( {\dfrac{D}{A}} \right) = \dfrac{2}{5}$ …….(1)
Therefore, $P\left( {\dfrac{D}{B}} \right)$ will be equal to the number of defective items in B divided by the total number of items in B.
So, $P\left( {\dfrac{D}{B}} \right) = \dfrac{1}{5}$ …….(2)
Now, we have picked up 3 items from A and 2 from B to put in C.
Therefore, when we pick up a random item in C, its proportion that it will be from A or B will be 3 : 2.
So, $P(A) = \dfrac{3}{5}$ and $P(B) = \dfrac{2}{5}$ ……….(3)
Now, we will use the formula given by the law of total probability which is given as follows:-
$P\left( {\dfrac{D}{C}} \right) = P(A).P\left( {\dfrac{D}{A}} \right) + P(B).P\left( {\dfrac{D}{B}} \right)$
where $P\left( {\dfrac{D}{C}} \right)$ represents the probability of getting a defective item randomly from slot C.
Now, using (1), (2) and (3) in this, we will have:-
$P\left( {\dfrac{D}{C}} \right) = \dfrac{3}{5} \times \dfrac{2}{5} + \dfrac{2}{5} \times \dfrac{1}{5}$
Simplifying it to get:-
$P\left( {\dfrac{D}{C}} \right) = \dfrac{6}{{25}} + \dfrac{2}{{25}}$
Hence, the answer is $P\left( {\dfrac{D}{C}} \right) = \dfrac{8}{{25}}$.

So, the correct answer is “Option C”.

Note: The students do not know the use of conditional probability and where do we tend to use it. If we have to find the probability of some event when some other thing has already happened which affects this like here, in our question, when we changed the slots, we got different probabilities of defective items. Therefore, we used conditional probability.