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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?

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.

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Important Questions from Probability

  1. Three dice are thrown. What is the probability of getting a sum which is a perfect square?

  2. Two distinct natural numbers from 1 to 9 are picked at random. What is the probability that their product has 1 in its unit place?

  3. Two dice are thrown. What is the probability that difference of numbers on them is 2 or 3 ?

  4. 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?

  5. What is the probability that all three boys sit together?

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