All Exams Test series for 1 year @ ₹349 only
Question

Consider the following for the next items that follow:

Three boys P, Q, R and three girls S, T, U are to be arranged in a row for a group photograph.

What is the probability that boys and girls sit alternatively?

This question was previously asked in
NDA I 2023 GAT Previous Year Paper (16-Apr-2023)
The correct answer is \(\frac{1}{10}\)

Understanding the Probability Problem

This problem asks for the probability of a specific arrangement when arranging 3 boys and 3 girls in a single row. The total number of individuals is 6 (3 boys + 3 girls).

We need to find the probability that the boys and girls sit in an alternating pattern. This means the arrangement must follow a sequence where a boy is followed by a girl, and a girl is followed by a boy.

Calculating Total Arrangements in a Row

First, let's determine the total number of ways to arrange these 6 distinct individuals (3 boys and 3 girls) in a row. The total number of permutations of n distinct items is given by n! (n factorial).

Here, we have 6 distinct individuals. So, the total number of arrangements is:

\( \text{Total arrangements} = 6! \)

Let's calculate 6!:

\( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \)

So, there are 720 total possible ways to arrange the 3 boys and 3 girls in a row.

Calculating Favorable Arrangements (Alternating Pattern)

Now, we need to find the number of arrangements where boys and girls sit alternatively. With 3 boys (B) and 3 girls (G), an alternating pattern in a row of 6 can only start with either a boy or a girl, and the sequence must perfectly alternate:

  • Pattern 1: Boy - Girl - Boy - Girl - Boy - Girl (B G B G B G)
  • Pattern 2: Girl - Boy - Girl - Boy - Girl - Boy (G B G B G B)

Let's consider Pattern 1 (B G B G B G):

  • The 3 boy positions are fixed (1st, 3rd, 5th). The 3 boys can be arranged in these 3 positions in \(3!\) ways.
  • The 3 girl positions are fixed (2nd, 4th, 6th). The 3 girls can be arranged in these 3 positions in \(3!\) ways.
  • The total number of arrangements for Pattern 1 is \(3! \times 3!\).

Let's calculate \(3!\):

\( 3! = 3 \times 2 \times 1 = 6 \)

So, the number of arrangements for Pattern 1 is \(6 \times 6 = 36\).

Now, let's consider Pattern 2 (G B G B G B):

  • The 3 girl positions are fixed (1st, 3rd, 5th). The 3 girls can be arranged in these 3 positions in \(3!\) ways.
  • The 3 boy positions are fixed (2nd, 4th, 6th). The 3 boys can be arranged in these 3 positions in \(3!\) ways.
  • The total number of arrangements for Pattern 2 is \(3! \times 3!\).

Again, \(3! \times 3! = 6 \times 6 = 36\).

The total number of favorable arrangements (where boys and girls sit alternatively) is the sum of arrangements from Pattern 1 and Pattern 2.

\( \text{Favorable arrangements} = (\text{Arrangements for BGBGBG}) + (\text{Arrangements for GBGBGB}) \)

\( \text{Favorable arrangements} = 36 + 36 = 72 \)

Calculating the Probability

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

\( \text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} \)

Using the values we calculated:

\( \text{Probability} = \frac{72}{720} \)

Now, we simplify the fraction:

\( \frac{72}{720} = \frac{72}{10 \times 72} = \frac{1}{10} \)

Thus, the probability that the boys and girls sit alternatively is \( \frac{1}{10} \).

Revision Table: Probability Concepts

Concept Description / Value
Total Individuals 3 Boys + 3 Girls = 6
Total Arrangements (6!) 720
Alternating Patterns BGBGBG or GBGBGB
Arrangements per Pattern (3! × 3!) \(6 \times 6 = 36\)
Total Favorable Arrangements \(36 + 36 = 72\)
Probability Formula \( \frac{\text{Favorable Arrangements}}{\text{Total Arrangements}} \)
Calculated Probability \( \frac{72}{720} = \frac{1}{10} \)

Additional Information on Permutations and Probability

This problem is a classic example of applying permutations in probability. A permutation is an arrangement of objects in a specific order, which is exactly what we deal with when arranging people in a row for a photograph.

The formula for the number of permutations of n distinct objects is \(n!\). When we have constraints, like the alternating seating pattern, we calculate the arrangements that meet those specific conditions and then use the probability formula.

Understanding factorial (\(n!\)) is crucial for permutations. It represents the product of all positive integers up to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

Probability problems involving arrangements often require identifying the total possible outcomes (total arrangements) and the specific outcomes that satisfy the given condition (favorable arrangements). The ratio of these two numbers gives the probability.

Was this answer helpful?

Similar Questions

  1. A coin is biased so that heads comes up thrice as likely as tails. In four independent tosses of the coin, what is probability of getting exactly three heads ?

  2. Suppose that there is a chance for a newly constructed building to collapse, whether the design is faulty or not. The chance that the design is faulty is 10%. The chance that the building collapses is 95% if the design is faulty, otherwise it is 45%. If it is seen that the building has collapsed, then what is the probability that it is due to faulty design?

  3. In a class, there are n students including the students P and Q. What is the probability that P and Q sit together if seats are assigned randomly?

  4. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on them is a prime number?

  5. A coin is biased so that heads comes up thrice as likely as tails. For three independent tosses of a coin, what is the probability of getting at most two tails?

  6. If the probability of simultaneous occurrence of two events A and B is p and the probability that exactly one of A, B occurs is q, then which of the following is/are correct/

    1) P(A̅) + P(B̅) = 2 – 2p – q

    2) P(A̅ ∩ B̅) = 1 – p – q

    Select the correct answer using the code given below:

  7. Consider the following in respect of two events A and B:

    1) P(A occurs but not B) = P(A) – P(B) if B ⊂ A

    2) P(A alone or B alone occurs) = P(A) + P(B) – P(A ∩ B)

    3) P(A ∪ B) = P(A) + P(B) if A and B are mutually exclusive

    Which of the above is/are correct?

  8. If A and B are two events such that 2P(A) = 3P(B), where 0 < P(A) < P(B) < 1, then which one of the following is correct?

  9. A box has ten chits numbered 0, 1, 2, 3, ……., 9. First, one chit is drawn at random and kept aside. From the remaining, a second chit is drawn at random. What is the probability that the second chit drawn is “9”?

  10. A bag contains 20 books out of which 5 are defective. If 3 of the books are selected at random and removed from the bag in succession without replacement, then what is the probability that all three books are defective?


Important Questions from Probability

  1. Two dice are thrown simultaneously. What is the probability of getting the same number on both the dice?

  2. The probability of being 53 Sundays in year 2020 is-

  3. Three dice are thrown randomly. The probability of coming 3 in at least one die is

  4. The probability of having 53 Tuesdays in an ordinary year is:

  5. When two dice are tossed simultaneously, the probability that the sum of the numbers appearing on both the dice is 8 will be

Need Expert Advice?
Upcoming Exams
NDA
September 13, 2026
CDS
September 13, 2026
Test Series
NDA img
Defence
NDA 2026 Mock Test Series (Latest Pattern)
501 Tests 1 Tests Free
844 Attempts
4.6(131)
English, Hindi

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App