Probability query

Question

8 Indians and 3 Americans are to stand in a row at random. What is the probability that no 2 Americans are always together?

I think the answer should be 1 because as there are many possible configurations, 2 Americans can never be "always" together.

However, a solution by a person online says the following :

so lets see.. it's given indians and americans so i'm assuming that implies every person is unique.

Our sample space would be $11! = 39916800$

Counting the unfavorable conditions (A-american I-indian) situation like this AAI(remaining in any order) AAA(remaining in any order)

AAI...... The number of ways to achieve this would be $3C2 *2* 8C1 * 8!$ which is choosing $2$ americans from $3$ which would be $3C2*2$ and choosing $1$ indian from $8$ >which would be 8C1 and then the remaining people in any order which would be 8!

AAA...... The number of ways to achieve this would be $3! * 8!$

So this would sum up to $2177280$

So the probability of this would be $2177280/39916800 = 0.0545$

So the probability of no $2$ americans being together would be $ 1 - 0.0545 = 0.9455$

so $94.55%$

Which one is the right answer?

Answer 1

For the case $AAA$ we have: $9\cdot 8!$ choices of seating.

For the case $AA$ we have: $ 8\cdot 7\cdot 8! + 2\cdot 8\cdot 8!$ choices of seating.

And there are a total of $11!$ choices of seating available. Thus $P(\text{no 2 A together}) = 1 - \dfrac{9\cdot 8! + 8\cdot 7\cdot 8! + 2\cdot 8\cdot 8!}{11!}$

Answer 2

8 Indians and 3 Americans are to stand in a row at random. What is the probability that no 2 Americans are always together?

This is simply badly phrased. Clearly you're after the proportion of all ways to arrange the row which do not place 2 Americans adjacent.

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My Approach

An equivalent problem is: How many ways can you put 8 balls into 4 boxes such that there is at least one ball in each of the middle two boxes? (The two end boxes can be empty.)

Where the 'balls' are Indians, and the 'boxes' are spaces between Americans. The order of individuals within the places doesn't matter.

This is a 'stars and bars' problem. There are ${n+k-1\choose k-1}$ ways to arrange $n$ objects into $k$ bins such that any bin may be empty. We require $n=6$ (because 2 of the 'balls' must be in the center bins, and we wish to count ways to arrange the other 6), and $k=4$.

This is: ${9\choose 3}$ ways to arrange places.

Now there are ${11\choose 3,8}$ ways to arrange the places in total.

$$\begin{align} \Pr(\text{answer})& =\frac{9!\,8!}{6!\,11!} \\[1ex] & = 0.5\dot{\overline{09}} \end{align}$$

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Your friend's approach

Our sample space would be 11!=39916800

The sample space is $11!/8!3!$, as we're only interested in arrangements of places, not individuals.

Counting the unfavorable conditions (A-american I-indian) situation like this AAI(remaining in any order) AAA(remaining in any order)

No, you don't want that. The first of these includes cases such as I...IAAAI...I, and does not include cases like I...IAI...IAA. The count will be off.

What you need to count is permutations of 8 I, 1 A and 1 AA, and then discount permutations of 8 I and 1 AAA. (Principle of Inclusion and Exclusion: count the ways at least one pair of Americans are adjacent, less ways all three are co-adjacent.)

$$\begin{align} \Pr(\text{answer}) & = 1 - \frac{{10\choose 8,1,1}-{9\choose 8,1}}{11\choose 8, 3} \\[1ex] & = 1 - \frac{\frac{10!}{8!\,1!\,1!}-\frac{9!}{8!\,1!}}{\frac{11!}{8!\,3!}} \\[1ex] & = \frac{11!-(10!-9!)3!}{11!} \\[1ex] & = 0.5\dot{\overline{09}} \end{align}$$