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Question

Consider the following data for the next three (03) items that follow :

There are 90 applicants for a job. Some of them are graduates. Some of them have less than three years experience.

Number of graduatesNumber of non-graduates
At least 3 years experience189
Less than 3 years experience3627

Let G be the event that the first applicant interviewed is a graduate and T be the event that first applicant interviewed has at least 3 years experience.

What is \(P(G | \overline T) \)  equal to?

This question was previously asked in
NDA I 2022 GAT Previous Year Paper (10-Apr-2022)
The correct answer is \(\frac{4}{7}\)

Calculating Conditional Probability \(P(G | \overline T)\) from Applicant Data

The problem asks us to calculate the conditional probability \(P(G | \overline T)\), which represents the probability that the first applicant interviewed is a graduate (\(G\)), given that the first applicant interviewed has less than three years of experience (\(\overline T\)). We are provided with data on 90 applicants categorized by their graduate status and years of experience.

Let's first organize the given data in a table:

At least 3 years experience (T) Less than 3 years experience (\(\overline T\)) Total
Number of graduates (G) 18 36 54
Number of non-graduates (\(\overline G\)) 9 27 36
Total 27 63 90

From the table, we can identify the number of applicants in different categories:

  • Number of graduates with at least 3 years experience: 18
  • Number of non-graduates with at least 3 years experience: 9
  • Number of graduates with less than 3 years experience: 36
  • Number of non-graduates with less than 3 years experience: 27
  • Total number of applicants: \(18 + 9 + 36 + 27 = 90\)

Understanding Conditional Probability \(P(A | B)\)

The formula for the conditional probability of event A occurring given that event B has occurred is:

\(P(A | B) = \frac{P(A \cap B)}{P(B)}\)

In our case, we want to find \(P(G | \overline T)\). This means we need to find the probability that an applicant is a graduate (\(G\)) given that they have less than 3 years experience (\(\overline T\)). Using the formula, we have:

\(P(G | \overline T) = \frac{P(G \cap \overline T)}{P(\overline T)}\)

Calculating \(P(G \cap \overline T)\) and \(P(\overline T)\)

First, let's find \(P(G \cap \overline T)\). This is the probability that an applicant is a graduate AND has less than 3 years experience. From the table, the number of applicants who are graduates and have less than 3 years experience is 36. The total number of applicants is 90.

So, \(P(G \cap \overline T) = \frac{\text{Number of graduates with < 3 years experience}}{\text{Total number of applicants}} = \frac{36}{90}\)

Next, let's find \(P(\overline T)\). This is the probability that an applicant has less than 3 years experience. From the table, the total number of applicants with less than 3 years experience is the sum of graduates with less than 3 years experience and non-graduates with less than 3 years experience, which is \(36 + 27 = 63\). The total number of applicants is 90.

So, \(P(\overline T) = \frac{\text{Total number of applicants with < 3 years experience}}{\text{Total number of applicants}} = \frac{63}{90}\)

Calculating \(P(G | \overline T)\)

Now we can substitute these probabilities into the conditional probability formula:

\(P(G | \overline T) = \frac{P(G \cap \overline T)}{P(\overline T)} = \frac{36/90}{63/90}\)

We can cancel out the denominator 90 from both the numerator and the denominator:

\(P(G | \overline T) = \frac{36}{63}\)

Simplifying the Probability Fraction

To simplify the fraction \(\frac{36}{63}\), we find the greatest common divisor (GCD) of 36 and 63. Both numbers are divisible by 9.

  • \(36 \div 9 = 4\)
  • \(63 \div 9 = 7\)

So, the simplified fraction is \(\frac{4}{7}\).

Therefore, \(P(G | \overline T) = \frac{4}{7}\).

Conclusion on Conditional Probability

The conditional probability \(P(G | \overline T)\), which is the probability that the first applicant interviewed is a graduate given they have less than 3 years experience, is equal to \(\frac{4}{7}\). This calculation is based directly on the data provided in the table for the 90 applicants.

Revision Table: Key Probability Concepts

Concept Definition Formula
Probability of Event A The likelihood of event A occurring. \(P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes}}\)
Joint Probability The probability of two events A and B both occurring. \(P(A \cap B)\)
Conditional Probability The probability of event A occurring given that event B has already occurred. \(P(A | B) = \frac{P(A \cap B)}{P(B)}\)
Complement of an Event The probability of event A not occurring. \(P(\overline A) = 1 - P(A)\)

Additional Information: Contingency Tables and Probability

The data presented in the problem is often called a contingency table or cross-tabulation table. These tables are very useful for calculating probabilities involving two categorical variables, like 'graduate status' and 'experience level' in this case. From a contingency table, we can easily calculate:

  • Marginal Probabilities: These are the probabilities of a single event occurring, found by dividing the row or column totals by the grand total. For example, \(P(G) = \frac{54}{90}\) (Total Graduates / Grand Total) or \(P(\overline T) = \frac{63}{90}\) (Total < 3 yrs experience / Grand Total).
  • Joint Probabilities: These are the probabilities of the intersection of two events, found by dividing the value in a cell by the grand total. For example, \(P(G \cap \overline T) = \frac{36}{90}\) (Graduates AND < 3 yrs experience / Grand Total).
  • Conditional Probabilities: These are probabilities where the sample space is restricted to a subset of the total outcomes. As shown in the solution, \(P(G | \overline T)\) focuses only on the row/column corresponding to \(\overline T\) and then finding the proportion of \(G\) within that subset. Another way to think about \(P(G | \overline T) = \frac{36}{63}\) is that *out of the 63 applicants who have less than 3 years experience*, 36 are graduates.

Understanding how to extract information from a contingency table is key to solving many probability problems, especially those involving conditional probability.

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Similar Questions

  1. What is \(P(\overline T | \overline G)\)  equal to?

  2. What is \(P (G \cap \overline T)\) equal to?

  3. Let two events A and B be such that P(A) = L and P(B) = M. Which one of the following is correct?

  4. If P(A|B) < P(A), then which one of the following is correct?

  5. If A and B are two events such that P(not A) = \(\rm \frac{7}{10}\) , P(not B) =  \(\rm \frac{3}{10}\)  and P(A|B) =  \(\rm \frac{3}{14}\) , then what is P(B|A) equal to?

  6. A and B are two events such that A̅ and B̅ are mutually exclusive. If P(A) = 0.5 and P(B) = 0.6, then what is the value of P(A|B)?

  7. Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys: 1 girl and 3 boys. One child is selected at random from each group. The probability that the three selected consist of 1 girl and 2 boys is

  8. Two integers x and y are chosen with replacement from the set (0, 1, 2…10). The probability that |x - y| > 5 is

  9. If two dice are thrown and at least one the dice show 5, then the probability that the sum is 10 or more is

  10. For two dependent events A and B, it is given that P(A) = 0.2 and P(B) = 0.5. If A ⊆ B, then the values of conditional probabilities P(A|B) and P(B|A) are respectively


Important Questions from Conditional Probability

  1. If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?

  2. Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) =  \(\dfrac{1}{4}\) and P(A̅) =  \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:

  3. A bike manufacturing factory has two plants P and Q. Plant P manufactures 60 percent of bikes and plant Q manufacture 40 percent. 80 percent of the bikes at plant P and 90 percent of the bikes at plant Q are rated of standard quality. A bike is chosen at random and is found to be of standard quality. What is the probability that it has come from plant P?

  4. 20 percent of the pens produced in a factory are of red colour and 4 percent are red and defective. If one pen is picked up at random, then what is the probability of its being defective if it is red?

  5. In a game, there are three rooms- I, Il and IIl. Room I contain 2 boxes having gift items and 3 empty boxes, room II contains 3 boxes having gift items and 2 empty boxes, and room III contains 4 boxes having gift items and one empty box respectively. There is an equal probability of each room being chosen by a player. Mr John selects one box from a room chosen at random. The probability that Mr John wins a box having gift items is:

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