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A group of individuals containing \(b\) boys and \(g\) girls is lined up in random order-that is, each of the \((b+g) !\) permutations is assumed to be equally likely. What is the probability that the person in the ith position, \(1 \leq i \leq\) \(b+g\), is a girl?

Short Answer

Expert verified
The probability that the person in the ith position, \(1 \leq i \leq b+g\), is a girl is \(\frac{1}{b+g}\).

Step by step solution

01

Find the total number of permutations

First, we need to find the total number of permutations without any specific conditions. This can be computed by simply calculating the factorial of the total number of individuals (boys and girls), which is \((b+g)!\).
02

Find permutations with a girl at the ith position

If we fix a girl at the ith position, we are left with \((b+(g-1))!\) permutations for the remaining individuals. Note that there are still \(g-1\) girls available.
03

Compute the probability

To compute the probability of having a girl at the ith position, we should divide the number of permutations with a girl at the ith position by the total number of permutations. Mathematically, we will have: Probability = \(\frac{Permutations\:with\:a\:girl\:at\:ith\:position}{Total\:number\:of\:permutations}\)
04

Plug in the values

Now, we plug in the values we computed in step 1 and 2: Probability = \(\frac{(b+(g-1))!}{(b+g)!}\)
05

Simplify the expression

Simplify the fraction: Probability = \(\frac{1}{b+g}\) The probability that the person in the ith position, \(1 \leq i \leq b+g\), is a girl is \(\frac{1}{b+g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Permutations
Imagine we have a group containing boys and girls. If we line them up randomly, each different order is called a permutation. The total permutations are all the different ways we can arrange these individuals. For a group of size \((b+g)\), there are \((b+g)!\) permutations. This is because for the first position, there are \(b+g\) options, for the second \((b+g-1)\), and so on.

What makes these permutations random is that each has an equal chance of occurring. This randomness assumes no specific arrangement is favored, such as all the boys or all the girls lining up first.
  • The larger the group, the more permutations you have.
  • Each possible line-up is equally likely.
  • Randomness is central to calculating probabilities in these scenarios.
Understanding random permutations is fundamental because it helps us calculate the likelihood of specific arrangements, like having a girl at a certain position.
Factorial Calculation
A factorial, represented by an exclamation mark \(!\), is a product of all positive integers up to a certain number. It's like multiplying a series of descending natural numbers.

For any number \(n\), the factorial \(n!\) is:

\(n! = n \times (n-1) \times (n-2) \times ... \times 1\)

For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials are used to calculate permutations because they represent all possible ways we can arrange items.
  • \((b+g)!\) gives the total arrangements of \(b+g\) people.
  • \((b+(g-1))!\) helps calculate scenarios when certain conditions are applied, like fixing a girl in a specific place.
Factorials are a foundational tool in probability and combinatorics. They prompt us to think about arranging and ordering objects, which is key in understanding permutations and combinations.
Conditional Probability
Conditional probability is about finding the chance of an event occurring given a certain condition. In our case, we want to know the probability that a specific position is occupied by a girl.

With random permutations, calculating this probability involves using the concept of factorials to count relevant arrangements. We calculate how many arrangements fit our condition (girl at the ith position) and divide it by the total number of arrangements.
The formula to find the probability of a given condition is expressed as: \[Probability = \frac{\text{Number of favorable permutations}}{\text{Total number of permutations}}\]
  • We use \((b+(g-1))!\) to determine ways to arrange the remaining people when a girl is fixed at the ith position.
  • Divide that number by \((b+g)!\) to get the probability.
  • The result is \(\frac{1}{b+g}\), meaning each person has an equal probability of being in any specific position.
Conditional probability lets us refine our calculations based on additional information we have, leading to more accurate predictions.

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Most popular questions from this chapter

In a state lottery, a player must choose 8 of the numbers from 1 to 40 . The lottery commission then performs an experiment that selects 8 of these 40 . numbers. Assuming that the choice of the lottery commission is equally likely to be any of the \(\left(\begin{array}{c}40 \\ 8\end{array}\right)\) combinations, what is the probability that a player has (a) all 8 of the numbers selected; (b) 7 of the numbers selected; (c) at least 6 of the numbers selected?

For any sequence of events \(E_{1}, E_{2}, \ldots\), define a new sequence \(F_{1}, F_{2}, \ldots\) of disjoint events (that is, events such that \(F_{i} F_{j}=\varnothing\) whenever \(i \neq j\) ) such that for all \(n \geq 1\), $$ \bigcup_{1}^{n} F_{i}=\bigcup_{1}^{n} E_{i} $$

There are \(n\) socks, 3 of which are red, in a drawer. What is the value of \(n\) if when 2 of the socks are chosen randomly, the probability that they are both red is \(\frac{1}{2} ?\)

Consider the matching problem, Example \(5 \mathrm{~m}\), and define \(A_{N}\) to be the number of ways in which the \(N\) men can select their hats so that no man selects his own. Argue that $$ A_{N}=(N-1)\left(A_{N-1}+A_{N-2}\right) $$ This formula, along with the boundary conditions \(A_{1}=0, A_{2}=1\), can then be solved for \(A_{N}\), and the desired probability of no matches would be \(A_{N} / N !\) HINT: After the first man selects a hat that is not his own, there remain \(N-1\) men to select among a set of \(N-1\) hats that does not contain the hat of one of these men. Thus there is one extra man and one extra hat. Argue that we can get no matches cither with the extra man selecting the extra hat or with the extra man not selecting the extra hat.

An urn contains \(n\) white and \(m\) black balls, where \(n\) and \(m\) are positive numbers. (a) If two balls are randomly withdrawn, what is the probability that they are the same color? (b) If a ball is randomly withdrawn and then replaced before the second one is drawn, what is the probability that the withdrawn balls are the same color? (c) Show that the probability in part (b) is always larger than the one in part (a).

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