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?
Calculating the Probability of Two Students Sitting Together
We are asked to find the probability that two specific students, P and Q, sit together in a class of \(n\) students when seats are assigned randomly.
To solve this, we need to determine two things:
- The total number of ways the \(n\) students can be seated.
- The number of ways P and Q can sit together.
Total Number of Seating Arrangements
If there are \(n\) students and \(n\) seats, the total number of ways to arrange these \(n\) students is the number of permutations of \(n\) distinct items, which is \(n!\) (n factorial).
Total number of outcomes = \(n!\)
Number of Arrangements Where P and Q Sit Together
To find the number of ways P and Q sit together, we can treat P and Q as a single unit or block. Now, instead of \(n\) individual students, we have (n-2) individual students plus the block containing P and Q. This gives us a total of (n-2) + 1 = (n-1) entities to arrange.
- The (n-1) entities can be arranged in \((n-1)!\) ways.
- Within the block containing P and Q, the two students can arrange themselves in \(2!\) ways (either P then Q, or Q then P). \(2! = 2\).
So, the total number of arrangements where P and Q sit together is the product of the number of ways to arrange the entities and the number of ways P and Q can arrange themselves within their block.
Number of favorable outcomes = \((n-1)! \times 2\)
Calculating the Probability
The probability that P and Q sit together is the ratio of the number of favorable outcomes (P and Q sit together) to the total number of possible outcomes (all possible seating arrangements).
Probability = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)
Probability = \(\frac{(n-1)! \times 2}{n!}\)
We know that \(n! = n \times (n-1)!\). Substituting this into the probability formula:
Probability = \(\frac{(n-1)! \times 2}{n \times (n-1)!}\)
We can cancel out the \((n-1)!\) term from the numerator and the denominator:
Probability = \(\frac{2}{n}\)
Therefore, the probability that P and Q sit together if seats are assigned randomly in a class of \(n\) students is \(\frac{2}{n}\).
Similar Questions
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Important Questions from Probability
- If one event is unaffected by the outcome of another event, the two events are said to be:
1. Dependent
2. Independent
3. Mutually exclusive
4. Mutually inclusive - Let $M$ be a randomly chosen non-empty subset of $S = \{1, 2, 3, \dots, 2026\}$.
Which of the following is the probability that the product of all the elements of $M$ is even? - A fair coin is tossed 6 times. The probability of getting at least 5 heads is _________________. (rounded off to two decimal places)
- Consider a random experiment of rolling a fair die. The probability of getting an odd number or a number less than 4 is ______ (Rounded off to two decimal places).
- In a null hypothesis significance test, the significance level $\alpha$ was set to 0.01. For which one or more of the following $p$ values would the null hypothesis be rejected?
