# Answer to Probability Question: Finding Outcomes Under Constraint

• Leo Liu
In summary: I'm not following. I followed the same idea and got ##4\cdot3\cdot2\quad 5\cdot4\cdot3\cdot\dbinom{2}{2}## as the numerator.In summary, there are ##8!## possible outcomes, where two of three end up in the same canoe.
Leo Liu
Homework Statement
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Relevant Equations
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The answer is ##4/7##. I know the denominator which is number of unconstrained outcomes is $$\dfrac{\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}}{4!}$$But I am not sure how to find the number of outcomes under the constraint. Could some give me a hint? Thanks.

Probabilities can often better be calculated if we consider the opposite event. How many possibilities are there, where two of three end up in the same canoe?

Leo Liu
For the complement described by @fresh_42, it's just worth to note that you first have to choose 2 of Barry, Carrie and Mary, then you need to choose which of the 4 boats these two people go into, then you need to assign the remaining 6 people to the other 3 boats.

[You can work either by imagining the boats are distinguishable or indistinguishable, although I think it's conceptually easier to treat them as distinguishable.]

Leo Liu
Note that it doesn't make any difference in which order you select the people to be randomly assigned to canoes. So, another approach is to imagine that the triplets are assigned first.

I got the following expression:
$$\xcancel{P(A)={1-\dfrac{\dbinom 3 2 \cdot 4\cdot \dfrac{\dbinom 6 2 \dbinom 4 2}{3!}}{\dfrac{\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}}{4!}}=\frac 4 7 }}$$
Thanks!

Last edited:
PeroK
Leo Liu said:
I got the following expression:
$$P(A)={1-\dfrac{\dbinom 3 2 \cdot 4\cdot \dfrac{\dbinom 6 2 \dbinom 4 2}{3!}}{\dfrac{\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}}{4!}}=\frac 4 7 }$$
Thanks!

I reckon it should be$$P(A) = 1 - \frac{ \dbinom 3 2 \cdot 4 \cdot \dbinom 6 2 \dbinom 4 2}{\dbinom 8 2 \dbinom 6 2 \dbinom 4 2} = \frac{4}{7}$$If you want, you could multiply by ##1 = (1/4!) / (1/4!)## to make it look like you're treating all the arrangements of the boats as indistinguishable.

PeroK and Leo Liu
It's a horrible way to do it in any case! If we put the triplets in first then the probability they all end up in different canoes is: $$p = 1 \times \frac 6 7 \times \frac 4 6 = \frac 4 7$$

etotheipi and Leo Liu
etotheipi said:
Yours is equal to ##-5/7##, I think.

See the note I added at the end of #6; it doesn't matter whether you take order into account for the boats. Doing so just amounts to dividing both the top and the bottom by ##4!##.
Sorry my bad. It appears bizarre to me that you get the probability even though you implicitly assign the order to boats by ##{\dbinom 8 2 \dbinom 6 2 \dbinom 4 2}##.

etotheipi said:
I reckon it should be$$P(A) = 1 - \frac{ \dbinom 3 2 \cdot 4 \cdot \dbinom 6 2 \dbinom 4 2}{\dbinom 8 2 \dbinom 6 2 \dbinom 4 2} = \frac{4}{7}$$If you want, you could multiply by ##1 = (1/4!) / (1/4!)## to make it look like you're treating all the arrangements of the boats as indistinguishable.
Or maybe I shouldn't multiply the pair on the same boat by 4 since all boats are indistinguishable from one another?
$$P(A)={1-\dfrac{\dbinom 3 2 \cdot \cancel 4\cdot \dfrac{\dbinom 6 2 \dbinom 4 2}{3!}}{\dfrac{\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}}{4!}}=\frac 4 7}$$

Leo Liu said:
Sorry my bad. It appears bizarre to me that you get the probability even though you implicitly assign the order to boats by ##{\dbinom 8 2 \dbinom 6 2 \dbinom 4 2}##.

Say you have a bag with 1 red, 1 blue and 1 green chip in it. What's the probability you get 1 red and 1 blue, if you choose 2 chips in succession without replacement?

- If you treat it with order, you win with RB or BR - (2 ways), out of a total possible BR, RB, BG, GB, RG, GR - (6 ways).

- If you treat it without order, you win with {RB} - (1 way), out of a total possible {RB}, {BG}, {RG} - (3 ways).

Either way, the probability is 1/3. If the 'permutation factor ##N!##' is the same in the numerator and denominator, it doesn't matter whether you take into account ordering.

Leo Liu
PeroK said:
It's a horrible way to do it in any case! If we put the triplets in first then the probability they all end up in different canoes is: $$p = 1 \times \frac 6 7 \times \frac 4 6 = \frac 4 7$$

I think you should really try to understand this solution.

Office_Shredder said:
I think you should really try to understand this solution.
I don't actually get. Mind explaining? I followed the same idea and got ##4\cdot3\cdot2\quad 5\cdot4\cdot3\cdot\dbinom{2}{2}## as the numerator.

You put the first of your three special people in any seat, and then you have 7 seats left but only 6 into which you can put the second special person, due to the constraint of the special people needing to be in different boats. After seating the second special person, you have 6 seats left but only 4 into which you can put the third special person.

Leo Liu
etotheipi said:
You put the first of your 3 special people in any seat, and then you have 7 seats left but only 6 into which you can put the second of your 3 special people, due to the constraint of the special people needing to be in different boats. After seating the second of your 3 special people, you have 6 seats left but only 4 into which you can put the third of your 3 special people.
This is smart. Thanks for your explanation.
Meh, maybe I am too dumb to understand combinatorics and probability.

Leo Liu said:
This is smart. Thanks for your explanation.
Meh, maybe I am too dumb to understand combinatorics and probability.

Don't thank me, it was @PeroK's idea!

Leo Liu said:
This is smart. Thanks for your explanation.
Meh, maybe I am too dumb to understand combinatorics and probability.
Sometimes it's good to count, but sometimes it's better to go directly for the probabilities.

Leo Liu
PS The important point is that it doesn't matter in what order you assign the seats randomly to the participants, there's the same probability of each outcome. That's what you need to think through and digest.

## 1. What is probability and how is it calculated?

Probability is a measure of the likelihood of an event occurring. It is calculated by dividing the number of desired outcomes by the total number of possible outcomes.

## 2. What is the difference between independent and dependent events in probability?

An independent event is one where the outcome of one event does not affect the outcome of another event. A dependent event is one where the outcome of one event is affected by the outcome of another event.

## 3. How do constraints affect the calculation of probabilities?

Constraints limit the possible outcomes and therefore impact the calculation of probabilities. They can reduce the total number of possible outcomes or change the likelihood of certain outcomes occurring.

## 4. What is the difference between a permutation and a combination in probability?

A permutation is an arrangement of a set of objects where order matters, while a combination is a selection of objects where order does not matter. Permutations involve all objects in a set, while combinations may involve only a subset of objects.

## 5. How can probability be used in real-life situations?

Probability can be used in real-life situations to make informed decisions, assess risks, and make predictions based on data. It is commonly used in fields such as finance, economics, and science to analyze and interpret data and make informed decisions.

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