Math Problem Statement

Suppose that 7 boys and 13 girls line up in a row. Let S be the number of places in the row where a boy and girl are standing next to each other. For example, for the row GBBGGGBGBGGGBGBGGBGG we have S = 12. Find the expected value of S (close nearest integer)

Solution

To find the expected value of SS, the number of places where a boy and a girl are standing next to each other in a line of 7 boys and 13 girls, we start by understanding the problem setup.

There are 20 positions in total, and we are interested in the number of adjacent pairs (i.e., consecutive pairs of positions) where one position is occupied by a boy and the other by a girl. There are n1n-1 pairs of adjacent positions in a row of nn positions. For n=20n = 20, there are 1919 adjacent pairs.

Next, consider the arrangement and the expected value. Let's denote the sequence of children as follows:

  • XiX_i is an indicator variable which is 1 if the ii-th and (i+1)(i+1)-th positions form a boy-girl pair or a girl-boy pair, and 0 otherwise.
  • S=i=119XiS = \sum_{i=1}^{19} X_i

To find the expected value of SS, we use the linearity of expectation: [ \mathbb{E}[S] = \mathbb{E}\left[\sum_{i=1}^{19} X_i\right] = \sum_{i=1}^{19} \mathbb{E}[X_i] ]

Each XiX_i is 1 if a boy and a girl (in either order) are next to each other at positions ii and i+1i+1.

The probability that the ii-th position is occupied by a boy is 720\frac{7}{20}, and the probability that it is occupied by a girl is 1320\frac{13}{20}.

Thus, the probability that positions ii and i+1i+1 are a boy-girl pair is: P(BiGi+1)=(720)(1319)\mathbb{P}(B_iG_{i+1}) = \left(\frac{7}{20}\right) \left(\frac{13}{19}\right) And similarly, the probability that positions ii and i+1i+1 are a girl-boy pair is: P(GiBi+1)=(1320)(719)\mathbb{P}(G_iB_{i+1}) = \left(\frac{13}{20}\right) \left(\frac{7}{19}\right)

Summing these two probabilities, we get: P(Xi=1)=(720)(1319)+(1320)(719)\mathbb{P}(X_i = 1) = \left(\frac{7}{20}\right) \left(\frac{13}{19}\right) + \left(\frac{13}{20}\right) \left(\frac{7}{19}\right) P(Xi=1)=91380+91380=182380=91190\mathbb{P}(X_i = 1) = \frac{91}{380} + \frac{91}{380} = \frac{182}{380} = \frac{91}{190}

The expected value of XiX_i is: E[Xi]=P(Xi=1)=91190\mathbb{E}[X_i] = \mathbb{P}(X_i = 1) = \frac{91}{190}

Since there are 19 such pairs, the expected value of SS is: E[S]=19×91190=17291909.1\mathbb{E}[S] = 19 \times \frac{91}{190} = \frac{1729}{190} \approx 9.1

Rounding to the nearest integer, the expected value of SS is: 9\boxed{9}

Would you like more details on any part of this solution, or do you have any other questions?

Here are five related questions you might find interesting:

  1. What is the probability that all boys are standing next to each other?
  2. How would the expected value of SS change if there were 10 boys and 10 girls?
  3. What is the expected number of boy-boy pairs in the row?
  4. If we know the exact sequence, how can we calculate SS directly?
  5. How would the expected value of SS change if the positions were arranged in a circle instead of a row?

Tip: When calculating expected values, breaking down the problem into smaller components and using the linearity of expectation can simplify the process significantly.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Linearity of Expectation

Formulas

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Theorems

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Suitable Grade Level

Advanced High School / College