Question

Lot A consists of 3G and 2D articles. Lot B consists of 4G and 1D articles. A new lot C is made by taking 3 articles from A and 2 articles from B. The probability of getting the article from C is
(a) \[\dfrac{1}{3}\]
(b) \[\dfrac{2}{5}\]
(c) \[\dfrac{8}{25}\]
(d) None

Answer 1

Hint: We solve this problem by taking the total probabilities of selecting 3 articles from A and 2 articles from B and we multiply them to the probability of getting one article from C because all the events are independent of each other but connect to the probability of getting one article from C.

Complete step-by-step solution
Let us assume that \[E\] be an event of selecting 3 articles from A
Since there are 3G and 2D articles in A we have three cases
Cases (i): getting 3G articles
The number of chances of getting 3G articles is \[{}^{3}{{C}_{3}}\].
Case (ii): getting 2G and 1D
Number of chances of getting 2G and 1D is \[{}^{3}{{C}_{2}}\times {}^{2}{{C}_{1}}\].
Case (iii) getting 1G and 2D
Number of chances of getting 1G and 2D is \[{}^{3}{{C}_{1}}\times {}^{2}{{C}_{2}}\].
Now let us find total number of possible chances as \[{}^{3}{{C}_{3}}+{}^{3}{{C}_{2}}\times {}^{2}{{C}_{1}}+{}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}=10\]
The total chances of selecting 3 articles from 5 articles of A is \[{}^{5}{{C}_{3}}=10\]
Now let us find the probability of happening of the event\[E\]is
\[\begin{align}
  & \Rightarrow P\left( E \right)=\dfrac{\text{total possible chances}}{\text{total chances}} \\
 & \Rightarrow P\left( E \right)=\dfrac{10}{10}=1 \\
\end{align}\]
Now let us assume that \[F\] be the event of selecting 2 articles from B
Since there are 4G and 1D articles in B we have three cases
Cases (i): getting 2G articles
Number of chances of getting 3G articles is \[{}^{4}{{C}_{2}}\].
Case (ii): getting 1G and 1D
Number of chances of getting 2G and 1D is \[{}^{4}{{C}_{1}}\times {}^{1}{{C}_{1}}\].
Now let us find total number of possible chances as \[{}^{4}{{C}_{2}}+{}^{4}{{C}_{1}}\times {}^{1}{{C}_{1}}=10\]
The total chances of selecting 2 articles from 5 articles of A is \[{}^{5}{{C}_{2}}=10\]
Now let us find the probability of happening of the event \[F\] is
\[\begin{align}
  & \Rightarrow P\left( F \right)=\dfrac{\text{total possible chances}}{\text{total chances}} \\
 & \Rightarrow P\left( F \right)=\dfrac{10}{10}=1 \\
\end{align}\]
Let us assume that \[H\] be event of getting 1 article from C.
We know that after selecting the 3 articles from A and 2 articles from B we have 5 articles in C.
We can write the probability of selecting one article from 5 articles in C is
\[\Rightarrow P\left( H \right)=\dfrac{1}{5}\]
Let us assume that T be the required probability.
Now we find the total probability of selecting one article from C at random can be calculated by using the formulae
\[\Rightarrow P\left( T \right)=P\left( E \right).P\left( F \right).P\left( H \right)\]
By substituting the values of all probabilities in above equation we get
\[\begin{align}
  & \Rightarrow P\left( T \right)=\left( 1 \right)\left( 1 \right)\left( \dfrac{1}{5} \right) \\
 & \Rightarrow P\left( T \right)=\dfrac{1}{5} \\
\end{align}\]
Therefore the answer is \[\dfrac{1}{5}\].

Note: we can solve this problem in another method. In the question, it is mentioned that 3 articles are selected from A and 2 articles from B. we need to find the probability of getting one article from 5 articles of C. as there is no need to get the different articles we can directly write
We can write the probability of selecting one article from 5 articles in C is
\[\Rightarrow P\left( H \right)=\dfrac{1}{5}\]
Therefore the answer is \[\dfrac{1}{5}\].