MHB A good fairy draws from the lottery wheel two successive balls

[mathmari]

Hey!

I am looking at the following and need some hints:

A lottery wheel contains five red, blue and yellow balls. For each color the balls are labeled with the numbers $ 1,2,3,4 $ and $ 5 $. A good fairy draws from the lottery wheel two successive balls. A ball which has already been drawn is not moved into the drum.

Do we consider here the hypergeometri distribution?

How can we model this experiment as a discrete probability space?

How could we calculate the probability that the ball that has been picked is labeled by $2$ ?

Also how could we calculate the probability that two balls with the label-sum $4$ are picked?

(Wondering)

 

[I like Serena]

Hey!

I am looking at the following and need some hints:

A lottery wheel contains five red, blue and yellow balls. For each color the balls are labeled with the numbers $ 1,2,3,4 $ and $ 5 $. A good fairy draws from the lottery wheel two successive balls. A ball which has already been drawn is not moved into the drum.

Do we consider here the hypergeometri distribution?

Hey mathmari! (Smile)

We would use a hypergeometric distribution if we'd look at a random variable that gives e.g. the number of green balls that we draw.
Is that the case?

How can we model this experiment as a discrete probability space?

A probability space is a triplet $(\Omega, \mathcal F, P)$.
What could each of those be? (Wondering)

How could we calculate the probability that the ball that has been picked is labeled by $2$ ?

I thought we were drawing 2 balls.
Which one is 'the' ball?

Also how could we calculate the probability that two balls with the label-sum $4$ are picked?

(Wondering)

Let's get back to that later. (Nerd)

 

[mathmari]

We would use a hypergeometric distribution if we'd look at a random variable that gives e.g. the number of green balls that we draw.
Is that the case?

No, we are just looking for the probability that we get a ball with a specific property, right? This is not a distribution that I mentionned in my other post (binomial, hypergeometric, geometric, poisson, exponential), is it? (Wondering)

A probability space is a triplet $(\Omega, \mathcal F, P)$.
What could each of those be? (Wondering)

Do we have that $\Omega$ is the set of all balls, $\mathcal F$ is the set of all the possible combinations of two balls and $P$ the probability of each event? (Wondering)

I thought we were drawing 2 balls.
Which one is 'the' ball?

We have three balls that have the label $2$, one in each colour. So, is the probability that the first ball has the label $2$ equal to $\frac{3}{15}$ and that the second ball has not the label $2$ the probability is $\frac{12}{15}$ ? Is the probability that the first ball has not the label $2$ equal to $\frac{12}{15}$ and that the second ball has the label $2$ the probability is $\frac{3}{14}$ ?
(Wondering)

 

[I like Serena]

No, we are just looking for the probability that we get a ball with a specific property, right? This is not a distribution that I mentionned in my other post (binomial, hypergeometric, geometric, poisson, exponential), is it? (Wondering)

I think we're looking at a tuple of 2 balls, each with specific properties, aren't we?
And indeed, that's yet another distribution.
Which one could it be?
What can we say if we compare the chances on specific tuples?

Do we have that $\Omega$ is the set of all balls, $\mathcal F$ is the set of all the possible combinations of two balls and $P$ the probability of each event? (Wondering)

A possible outcome would be that we first draw the green 2, and then draw the red 5.
So that should be in the sample space shouldn't it?
A possible event is that we draw 2 balls that have a label-sum of 4, which would be a set of tuples that satisfy the condition.

We have three balls that have the label $2$, one in each colour. So, is the probability that the first ball has the label $2$ equal to $\frac{3}{15}$ and that the second ball has not the label $2$ the probability is $\frac{12}{15}$ ? Is the probability that the first ball has not the label $2$ equal to $\frac{12}{15}$ and that the second ball has the label $2$ the probability is $\frac{3}{14}$ ?
(Wondering)

All correct, except that after we have drawn the first ball, there are 14 balls left.
So the probability that the first ball has the label $2$ is indeed equal to $\frac{3}{15}$, but that the second ball has not the label $2$ has probability $\frac{12}{14}$.

But it's still not clear to me what it means that 'the' ball is a 2.
Is it intended that at least one of the two balls is a 2? (Wondering)

 

[mathmari]

I think we're looking at a tuple of 2 balls, each with specific properties, aren't we?
And indeed, that's yet another distribution.
Which one could it be?
What can we say if we compare the chances on specific tuples?

I don't really know what it could be. (Wondering)

All correct, except that after we have drawn the first ball, there are 14 balls left.
So the probability that the first ball has the label $2$ is indeed equal to $\frac{3}{15}$, but that the second ball has not the label $2$ has probability $\frac{12}{14}$.

Oh yes (Blush)

But it's still not clear to me what it means that 'the' ball is a 2.
Is it intended that at least one of the two balls is a 2? (Wondering)

Oh sorry! It should be "the probability that one of the picked ball is labeled by $2$". (Tmi)

So is this the sum of the probability that the first one is labeled by $2$ and the probability that the second one is labeled by $2$ ? (Wondering)

 

[I like Serena]

I don't really know what it could be. (Wondering)

What is the chance that we draw (red 2, blue 5)?
And what is the chance that we draw (green 1, blue 3)?

Oh sorry! It should be "the probability that one of the picked ball is labeled by $2$". (Tmi)

So is this the sum of the probability that the first one is labeled by $2$ and the probability that the second one is labeled by $2$ ? (Wondering)

That would only be true if those events were mutually exclusive.
But they are not. It's possible for both balls to be labeled 2.
It means that we have to apply the general sum rule.

For reference, the sum rule for mutually exclusive events A and B is: $P(A\cup B)=P(A)+P(B)$.
And the general sum rule for any events A and B is: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

 

[mathmari]

What is the chance that we draw (red 2, blue 5)?
And what is the chance that we draw (green 1, blue 3)?

Do we use here the binomial coefficient? (Wondering)

That would only be true if those events were mutually exclusive.
But they are not. It's possible for both balls to be labeled 2.
It means that we have to apply the general sum rule.

For reference, the sum rule for mutually exclusive events A and B is: $P(A\cup B)=P(A)+P(B)$.
And the general sum rule for any events A and B is: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

Ah ok!

So we have that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{3}{15}\cdot \frac{12}{14}+\frac{12}{15}\cdot \frac{3}{14}-\frac{3}{15}\cdot \frac{2}{14}$$ right? (Wondering)

 

[mathmari]

Also how could we calculate the probability that two balls with the label-sum $4$ are picked?

Do we have to calculate the probability $$P((\text{ Ball with label } 1) \cup (\text{ Ball with label } 3))+P((\text{ Ball with label } 3) \cup (\text{ Ball with label } 1))+P((\text{ Ball with label } 2) \cup (\text{ Ball with label } 2))$$ ? (Wondering)

We have that $P((\text{ Ball with label } 1) \cup (\text{ Ball with label } 3))=\frac{3}{15}\cdot \frac{3}{14}$, $P((\text{ Ball with label } 3) \cup (\text{ Ball with label } 1))=\frac{3}{15}\cdot \frac{3}{14}$ and $P((\text{ Ball with label } 2) \cup (\text{ Ball with label } 2))=\frac{3}{15}\cdot \frac{2}{14}$, or not?

Therefore we get $$P(\text{ two balls with the label-sum } 4 \text{ are picked })=\frac{3}{15}\cdot \frac{3}{14}+\frac{3}{15}\cdot \frac{3}{14}+\frac{3}{15}\cdot \frac{2}{14}$$

Is that correct? (Wondering)

 

[I like Serena]

Do we use here the binomial coefficient? (Wondering)

No. The chance that the first ball is red and labeled 2 is $\frac 1{15}$.
The chance that the second ball is blue and labeled 5 is $\frac 1{14}$, since there are 14 balls left.
So:
$$P(\text{red 2, blue 5}) = \frac 1{15}\cdot\frac 1{14}$$
In the same way, we have:
$$P(\text{green 1, blue 3}) = P(\text{red 2, blue 5}) = \frac 1{15}\cdot\frac 1{14}$$
In other words, the probability on any outcome, consisting of 2 different balls drawn, is the same.

Do we know of a distribution where all probabilities are the same? (Wondering)

(Btw, the notation (x,y) is also known as a tuple or a 2-tuple or an ordered pair.)

Ah ok!

So we have that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{3}{15}\cdot \frac{12}{14}+\frac{12}{15}\cdot \frac{3}{14}-\frac{3}{15}\cdot \frac{2}{14}$$ right? (Wondering)

Yep. (Nod)

Do we have to calculate the probability $$P((\text{ Ball with label } 1) \cup (\text{ Ball with label } 3))+P((\text{ Ball with label } 3) \cup (\text{ Ball with label } 1))+P((\text{ Ball with label } 2) \cup (\text{ Ball with label } 2))$$ ? (Wondering)

We have that $P((\text{ Ball with label } 1) \cup (\text{ Ball with label } 3))=\frac{3}{15}\cdot \frac{3}{14}$, $P((\text{ Ball with label } 3) \cup (\text{ Ball with label } 1))=\frac{3}{15}\cdot \frac{3}{14}$ and $P((\text{ Ball with label } 2) \cup (\text{ Ball with label } 2))=\frac{3}{15}\cdot \frac{2}{14}$, or not?

Therefore we get $$P(\text{ two balls with the label-sum } 4 \text{ are picked })=\frac{3}{15}\cdot \frac{3}{14}+\frac{3}{15}\cdot \frac{3}{14}+\frac{3}{15}\cdot \frac{2}{14}$$

Is that correct? (Wondering)

Not quite.

Note that $P((\text{ Ball with label } 2) \cup (\text{ Ball with label } 2))$ is the probability that at least one ball is labeled 2 AND at least one ball is labeled 2.
That is, we include the possibility that the first ball is 2 while the second ball is NOT 2.
That can't be right can it? (Worried)

Instead we should find the probability:
$$P(\text{first ball is labeled $x$} \land \text{second ball is labeled $y$} \land x+y=4)
=P((1,3) \cup (3,1) \cup (2,2))
$$
where $(x,y)$ denotes the set of outcomes with the first ball labeled $x$ and the second ball labeled $y$.
(Thinking)