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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. 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)?

  2. 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

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

  4. If A and B are two events such that P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.4, then consider the following statements:

    1. P(A̅ ∪ B) = 0.9

    2. P(B̅ | A̅) = 0.6

    Which of the statements is / are correct? 

  5. For two mutually exclusive events A and B, P(A) = 0.2 and P (A̅ ∩ B) = 0.3. What is P (A|(A ∪ B)) equal to?

  6. Consider the following statements:

    1. If A and B are mutually exclusive events, then it is possible that P(A) = P(B) = 0.6.

    2. If A and B are any two events such that P(A|B) = 1, then P(B̅|A̅) = 1

    Which of the above statement is/are correct?
  7. If A, B, C are three events, then what is the probability that at least two of these events occur together?

  8. If 5 of a Company’s 10 delivery trucks do not meet emission standards and 3 of them are chosen for inspection, then what is the probability that none of the trucks chosen will meet emission standards?

  9. A problem is given to three students A, B and C whose probabilities of solving the problem are \(\frac{1}{2},\frac{3}{4}\) and \(\frac{1}{4}\)  respectively. What is the probability that the problem will be solved if they all solve the problem independently?

  10. 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


Important Questions from Conditional Probability

  1. 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)?

  2. 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

  3. In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?

  4. Let X1 and X2 be independent random variables each having geometric distribution qk p ; k = 0, 1, 2, …. Then the conditional distribution of X1 given X1 + X2 is

  5. Find the probability that they speak the same fact if A speaks truth in 60% cases and B speaks truth in 75% cases?

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