For an oral test ‘25’ questions are prepared in which 5 questions are good and ‘20’ questions are difficult. If two questions are given to two candidates A and B in the order, the probability that ‘B’ gets a good question is
\[\left( \text{a} \right)\text{ }\dfrac{4}{5}\]
\[\left( \text{b} \right)\text{ }\dfrac{1}{5}\]
\[\left( \text{c} \right)\text{ }\dfrac{1}{6}\]
\[\left( \text{d} \right)\text{ }\dfrac{5}{6}\]

Hint: To solve the given question, we will first find out what the definition of probability is. Then we will form two cases according to the question. In case I, we will find the probability that A and B both get a good question and in case II, we will find the probability that A gets a difficult question while B gets a good question. Then we will add both the probabilities to get the total probability that B will get a good question.

Complete step-by-step answer:
Before we start to solve the above question, we must first know what probability is. Probability is defined as the chances or possibilities of happening at any random event. It is denoted by P(E). The probability is given by the formula.
\[P\left( E \right)=\dfrac{\text{Favourable Outcomes}}{\text{Total number of Outcomes}}\]
Now, it is given that the first question is given to A and then the next question is given to B. Thus, there are two possibilities: either A gets a good question or a difficult question. Thus, we are going to form two cases on the basis of this.
Case I: ‘A’ gets a good question.
There are 5 good questions and ‘A’ gets one of them. Now, ‘B’ also has to get a good question so he will have to get one from the remaining good question. Let \[P\left( {{A}_{G}} \right)\] be the probability that ‘A’ gets a good question and \[P\left( {{B}_{G}} \right)\] is the probability that ‘B’ gets a good question. Thus, we have,
\[P\left( {{B}_{1}} \right)=P\left( {{A}_{G}} \right).P\left( {{B}_{G}} \right)\]
\[\Rightarrow P\left( {{B}_{1}} \right)=\dfrac{5}{25}\times \dfrac{4}{24}\]
\[\Rightarrow P\left( {{B}_{1}} \right)=\dfrac{20}{25\times 24}\]
Case II: ‘A’ gets a difficult question.
There are 20 difficult questions and ‘A’ gets one of them. Now, ‘B’ will get a good question. Let \[P\left( {{A}_{D}} \right)\] be the probability that ‘A’ gets a difficult question. Thus, we have,
\[P\left( {{B}_{2}} \right)=P\left( {{A}_{D}} \right).P\left( {{B}_{G}} \right)\]
\[\Rightarrow P\left( {{B}_{2}} \right)=\dfrac{20}{25}\times \dfrac{5}{24}\]
\[\Rightarrow P\left( {{B}_{2}} \right)=\dfrac{100}{25\times 24}\]
Now, the total probability will be the sum of \[P\left( {{B}_{1}} \right)\] and \[P\left( {{B}_{2}} \right).\] Thus, we will get,
\[\text{Total Probability}=P\left( {{B}_{1}} \right)+P\left( {{B}_{2}} \right)\]
\[\Rightarrow \text{Total Probability}=\dfrac{20}{25\times 24}+\dfrac{100}{25\times 24}\]
\[\Rightarrow \text{Total Probability}=\dfrac{120}{25\times 24}\]
\[\Rightarrow \text{Total Probability}=\dfrac{1}{5}\]
Hence, option (b) is the right answer.

Note: We cannot solve the question as shown below. The favorable outcomes for B will be 5 and the total number of outcomes will be 25. Thus, the probability will be
\[P\left( B \right)=\dfrac{5}{25}\]
\[\Rightarrow P\left( B \right)=\dfrac{1}{5}\]
This is the wrong method as P(B) is not independent. It is dependent on the probability of A also.