Question

There are four machines and it is known that exactly two of them are faulty. They are tested one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is
A. \[\dfrac{1}{3}\]
B. \[\dfrac{1}{6}\]
C. \[\dfrac{1}{2}\]
D. \[\dfrac{1}{4}\]

Answer 1

Hint: Recall the concepts of probability and the formula to find the probability of an event. We are asked to find the probability to find out the faulty ones in just two tests. Check for all the possible cases, find the probability of each case and combine them to find the required probability.

Complete step-by-step answer:
Given, total number of machines \[T = 4\]
Number of faulty machines \[F = 2\]
Total number of tests, \[S = 2\]
Let us find the number of working machines.
Number of working machines will be,
 \[W = T - F\]
Putting the values of \[T\] and \[F\] we get
 \[W = 4 - 2 \\
   \Rightarrow W = 2 \]
Total number of tests is given as two and we have two working machines and two faulty ones. In order to identify both the faulty machines in two tests, either both the machines should be the working ones or the faulty ones.
For probability of an event we have the formula,
 \[P(E) = \dfrac{{{\text{Favourable number of events}}}}{{{\text{total number of events}}}}\]
There are be two cases
Case 1: both the machines are working ones
For the first test favourable number of events will be two, as there are two working machines and in total are four machines, so probability will be
 \[P({W_1}) = \dfrac{2}{4}\]
For the second test the remaining number of working machine will be one and total machines now will be three, so probability will be
 \[P({W_2}) = \dfrac{1}{3}\]
Total probability for case 1 is
 \[P(W) = \dfrac{2}{4} \times \dfrac{1}{3} = \dfrac{1}{6}\]
Case 2: both the machines are faulty ones
For the first test favourable number of events will be two, as there are two faulty machines and in total are four machines, so probability will be
 \[P({F_1}) = \dfrac{2}{4}\]
For the second test the remaining number of faulty machine will be one and total machines now will be three, so probability will be
 \[P({F_2}) = \dfrac{1}{3}\]
Total probability for case 1 is
 \[P(F) = \dfrac{2}{4} \times \dfrac{1}{3} = \dfrac{1}{6}\]
Now, total probability will be,
 \[P = P(W) + P(F)\]
Putting the values of \[P(W)\] and \[P(F)\] , we have
 \[P = \dfrac{1}{6} + \dfrac{1}{6} \\
   \Rightarrow P = \dfrac{1}{3} \]
Therefore probability that only two test will be needed is \[\dfrac{1}{3}\]
So, the correct answer is “Option A”.

Note: There are few concepts of probability that one should always remember. If the possibility of an event be of two cases let’s say case A or case B then the total probability is obtained by adding the probabilities of both the cases but if possibility of an event is case A and case B then total probability is obtained by multiplying the probabilities of both the cases.