What is the probability that Ugo and Zoe will not meet at the party?

[DottZakapa]

Homework Statement

Aldo has 6 friends, of whom 3 will be invited to a party. If all the possible choices are equally likely, what is the probability that Ugo and Zoe (2 of his friends) will not meet at the party?

Relevant Equations

Probability

So first i compute the number of manners i can choose a sett of 3 ppl among 6

$\frac {something}{\binom {6}{3}}$

but then i don't understand from where should i pick up the two friends, from the set of those going to the party or what?
Sorry but have problems on understanding how to deal with such kind of problems

 

[PeroK]

Sorry but have problems on understanding how to deal with such kind of problems

There are only six people involved, so why not write out all the possibilities?

 

[etotheipi]

How many ways can you choose 3 people from the group of 6, such that 2 of them are fixed to be Ugo and Zoe? Then, can you write down the probability that Ugo and Zoe do meet? How do you get the probability that Ugo and Zoe don't meet from that?

 

[DottZakapa]

There are only six people involved, so why not write out all the possibilities?

so if I choose 2 guy from the six, all the possible manners would be:

$\binom {6}{2}$ is this what are you suggesting?

 

[PeroK]

so if I choose 2 guy from the six, all the possible manners would be:

$\binom {6}{2}$ is this what are you suggesting?

Assume his friends are $A, B, C, D, U, Z$. You have $\binom 6 3 = 20$ ways to pick the three party guests (as you have already calculated). Write these all down:

ABC, ABD, ABU, ABZ, ACD, ... DUZ

And count how many have both U and Z at the party and how many don't. This is good practice.

 

[DottZakapa]

How many ways can you choose 3 people from the group of 6, such that 2 of them are fixed to be Ugo and Zoe?

$\binom{5}{3}$ ?

Then, can you write down the probability that Ugo and Zoe do meet?

no

 

[DottZakapa]

Assume his friends are $A, B, C, D, U, Z$. You have $\binom 6 3 = 20$ ways to pick the three party guests (as you have already calculated). Write these all down:

ABC, ABD, ABU, ABZ, ACD, ... DUZ

And count how many have both U and Z at the party and how many don't. This is good practice.

now i try

 

[DottZakapa]

Assume his friends are $A, B, C, D, U, Z$. You have $\binom 6 3 = 20$ ways to pick the three party guests (as you have already calculated). Write these all down:

ABC, ABD, ABU, ABZ, ACD, ... DUZ

And count how many have both U and Z at the party and how many don't. This is good practice.

so i have done that and in the 20 sets, 16 are sets in which U and Z are not together . now how do I translate it into binomial?

 

[PeroK]

so i have done that and in the 20 sets, 16 are sets in which U and Z are not together . now how do I translate it into binomial?

In this case it is easier to calculate the complementary probability that U and Z are together. That's only 4 options. Can you see a quick way to calculate that? Or, translate that into a binomial?

 

[PeroK]

Hint: write down the four cases where U and Z meet at the party:

AUZ, BUZ, CUZ, DUZ

What do you notice about this? Why are there four cases here?

 

[DottZakapa]

In this case it is easier to calculate the complementary probability that U and Z are together. That's only 4 options. Can you see a quick way to calculate that? Or, translate that into a binomial?

I've computed the correct solution, that is 4/5 but i want understand haw to get that there are 4 set and 16 not drawing the set. without drawing how should I reason in order to come up with a binomial?

 

[PeroK]

I've computed the correct solution, that is 4/5 but i want understand haw to get that there are 4 set and 16 not drawing the set. without drawing how should I reason in order to come up with a binomial?

See my last post above.

 

[DottZakapa]

Hint: write down the four cases where U and Z meet at the party:

AUZ, BUZ, CUZ, DUZ

What do you notice about this? Why are there four cases here?

on each set i have 3! combinations

 

[PeroK]

on each set i have 3! combinations

Let me explain:

We want to calculate the number of ways that U and Z can meet at the party. Both U and Z must be there. That leaves only one space for one more friend. There are four more friends so $\binom 4 1 = 4$ ways to choose that friend.

That's the simplest way: if U and Z meet at the party then the guests are U, Z and one more guest from a choice of four.

 

[DottZakapa]

Let me explain:

We want to calculate the number of ways that U and Z can meet at the party. Both U and Z must be there. That leaves only one space for one more friend. There are four more friends so $\binom 4 1 = 4$ ways to choose that friend.

That's the simplest way: if U and Z meet at the party then the guests are U, Z and one more guest from a choice of four.

With the previous example of the car I pictured how to proceed, but also with this. From this probability then I just subtract from 1 in order to get the probability of not meeting .
Ok, thanks for the explanation and patient.