What is the Probability of Selecting a White Ball from Two Boxes?

[mathmari]

Hey!

I am looking at the following:

We have two boxes. In the blue box there are one black and three white balls. In the red box there are two black and four white balls. We choose one of the boxes and then we pick a ball from that box.

So we have that $\Omega=\{B,R\}\times \{b,w\}$, where B:blue, R:red, b:black, w:white.

We have that $P(B)=P(R)=0,5$ since we have two boxes and the probability to take one of them is equivalent, right? (Wondering)

To calculate the probabilities $P(w\mid B), P(b\mid B), P(w\mid R), P(b\mid R)$ do we use the Bayes’ theorem?

$$P(w\mid B)=\frac{P(w\cap B)}{P(B)}$$ Are the events w and B independent? If yes, we get the following: $$P(w\mid B)=\frac{P(w)\cdot P( B)}{P(B)}=P(w)$$ Is that correct (Wondering) But in that way we don't take into consideration from which box we take the ball, do we? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

Hey!

I am looking at the following:

We have two boxes. In the blue box there are one black and three white balls. In the red box there are two black and four white balls. We choose one of the boxes and then we pick a ball from that box.

So we have that $\Omega=\{B,R\}\times \{b,w\}$, where B:blue, R:red, b:black, w:white.

We have that $P(B)=P(R)=0,5$ since we have two boxes and the probability to take one of them is equivalent, right? (Wondering)

It doesn't say that we randomly pick a box does it? So I don't think we can assume that.
Perhaps we don't need it. (Thinking)

To calculate the probabilities $P(w\mid B), P(b\mid B), P(w\mid R), P(b\mid R)$ do we use the Bayes’ theorem?
$$P(w\mid B)=\frac{P(w\cap B)}{P(B)}$$

I'm afraid that that's not Bayes' theorem. Instead it's the definition of a conditional probability.

For reference Bayes' theorem is:
$$P(A\mid B) = \frac{P(B \mid A) \, P(A)}{P(B)}$$
It converts a conditional probability into an inverted conditional probability

Are the events w and B independent? If yes, we get the following: $$P(w\mid B)=\frac{P(w)\cdot P( B)}{P(B)}=P(w)$$ Is that correct (Wondering) But in that way we don't take into consideration from which box we take the ball, do we? (Wondering)

We can already see that those events won't be independent.
If we pick box B, the chance that we pick a white ball is different from when we'd pick box R.

Let's take a look at what $P(w\mid B)$ actually means.
It is the probability that we draw a white ball given that we have picked box B.
What is the probability to draw a white ball if we assume we only have one box, which is box B? (Wondering)

 

[mathmari]

It doesn't say that we randomly pick a box does it? So I don't think we can assume that.
Perhaps we don't need it. (Thinking)

I read it again. It is mentionned that we randomly pick a box.

I'm afraid that that's not Bayes' theorem. Instead it's the definition of a conditional probability.

For reference Bayes' theorem is:
$$P(A\mid B) = \frac{P(B \mid A) \, P(A)}{P(B)}$$
It converts a conditional probability into an inverted conditional probability
We can already see that those events won't be independent.
If we pick box B, the chance that we pick a white ball is different from when we'd pick box R.

Let's take a look at what $P(w\mid B)$ actually means.
It is the probability that we draw a white ball given that we have picked box B.
What is the probability to draw a white ball if we assume we only have one box, which is box B? (Wondering)

Do we have to divide the number of white balls in B by the total number of balls in B? (Wondering)

So we don't have to use the conditional probability? (Wondering)

 

[I like Serena]

I read it again. It is mentionned that we randomly pick a box.

Doesn't it say that we 'choose' a box?
That doesn"t sound like selecting randomly and aselectively. (Wondering)

Do we have to divide the number of white balls in B by the total number of balls in B? (Wondering)

So we don't have to use the conditional probability? (Wondering)

Yep. (Nod)

 

[mathmari]

Doesn't it say that we 'choose' a box?
That doesn"t sound like selecting randomly and aselectively. (Wondering)

It says that "We randomly choose one of the two boxes and then we pick a ball from that box." Is the probability that we pick a black ball equal to the sum of $P(b\mid B)$ and $P(b\mid R)$ ? (Wondering)

We assume that a ball is white. Is the probability that this ball is from the red box equal to $P(w\mid R)$ ? (Wondering)

 

[I like Serena]

It says that "We randomly choose one of the two boxes and then we pick a ball from that box."

Ah okay. That was missing from post #1 then.

Is the probability that we pick a black ball equal to the sum of $P(b\mid B)$ and $P(b\mid R)$ ? (Wondering)

Let's see... suppose there were 99 black balls out of a 100 in both B and R.
Then:
$$P(b\mid B) + P(b\mid R) = \frac{99}{100} + \frac{99}{100} = \frac{198}{100}$$
A probability cannot be greater than $1$ can it? (Wondering)

Which calculation rules for probabilities do you know?
Can we apply them?

We assume that a ball is white. Is the probability that this ball is from the red box equal to $P(w\mid R)$ ? (Wondering)

$P(w\mid R)$ is the probability that a ball is white, given that it's coming from R.
That doesn't sound like the same thing does it? (Wondering)

 

[mathmari]

Let's see... suppose there were 99 black balls out of a 100 in both B and R.
Then:
$$P(b\mid B) + P(b\mid R) = \frac{99}{100} + \frac{99}{100} = \frac{198}{100}$$
A probability cannot be greater than $1$ can it? (Wondering)

Which calculation rules for probabilities do you know?
Can we apply them?

Could you give me a hint? I got stuck right now... (Thinking)

$P(w\mid R)$ is the probability that a ball is white, given that it's coming from R.
That doesn't sound like the same thing does it? (Wondering)

Do we maybe have to calculate the probability $P(R\mid w)$ ? This is the probability that, given that we have chosen a white ball, it is from the red box, or not? (Wondering)

 

[I like Serena]

Could you give me a hint? I got stuck right now... (Thinking)

Of course.
We have:
$$P(b) = P((b\land R)\lor(b\land B))$$
That is, either we pick a black ball from the red box, or we pick a black ball from the blue box.
Since picking from either R or B is mutually exclusive, we can use the sum rule to conclude that:
$$ P((b\land R)\lor(b\land B))=P(b\land R)+P(b\land B)$$
And from the definition of the conditional probability it follows (by the general product rule) that:
$$P(b\land R)+P(b\land B)=P(b\mid R)P(R) + P(b\mid B)P(B)$$
Can we find the probability now? (Wondering)

Do we maybe have to calculate the probability $P(R\mid w)$ ? This is the probability that, given that we have chosen a white ball, it is from the red box, or not? (Wondering)

Yes. This is where Bayes' theorem would come in. (Thinking)

 

[mathmari]

Of course.
We have:
$$P(b) = P((b\land R)\lor(b\land B))$$
That is, either we pick a black ball from the red box, or we pick a black ball from the blue box.
Since picking from either R or B is mutually exclusive, we can use the sum rule to conclude that:
$$ P((b\land R)\lor(b\land B))=P(b\land R)+P(b\land B)$$
And from the definition of the conditional probability it follows (by the general product rule) that:
$$P(b\land R)+P(b\land B)=P(b\mid R)P(R) + P(b\mid B)P(B)$$
Can we find the probability now? (Wondering)

Ah I see!

So we have the following: $$P(b)=P(b\mid R)P(R) + P(b\mid B)P(B)=\frac{1}{3}\cdot 0,5 + \frac{1}{4}\cdot 0,5=\frac{1}{6}+\frac{1}{8}=\frac{14}{48}=\frac{7}{24}$$ right? (Wondering)

Yes. This is where Bayes' theorem would come in. (Thinking)

Ah ok! So, we have the following: $$P(R\mid w)=\frac{P(R\cap w)}{P(w)}=\frac{P(w\mid R)\cdot P(R)}{P(w)}=\frac{\frac{2}{3}\cdot 0,5}{P(w)}=\frac{\frac{2}{6}}{P(w)}=\frac{1}{3\cdot P(w)}$$ where $$P(w)=P((w\land R)\lor(w\land B))=P(w\land R)+P(w\land B)=P(w\mid R)P(R) + P(w\mid B)P(B)=\frac{2}{3}\cdot 0,5 + \frac{3}{4}\cdot 0,5=\frac{2}{6} + \frac{3}{8}=\frac{17}{24}$$
Therefore, we get $$P(R\mid w)=\frac{1}{3\cdot \frac{17}{24}}=\frac{1}{\frac{17}{8}}=\frac{8}{17}$$
Is everything correct? (Wondering) Why do we use the Bayes' theorem to calculate $P(R\mid w)$ but not for $P(w\mid R)$ ? (Wondering)

 

[I like Serena]

Ah I see!

So we have the following: $$P(b)=P(b\mid R)P(R) + P(b\mid B)P(B)=\frac{1}{3}\cdot 0,5 + \frac{1}{4}\cdot 0,5=\frac{1}{6}+\frac{1}{8}=\frac{14}{48}=\frac{7}{24}$$ right? (Wondering)

Ah ok! So, we have the following: $$P(R\mid w)=\frac{P(R\cap w)}{P(w)}=\frac{P(w\mid R)\cdot P(R)}{P(w)}=\frac{\frac{2}{3}\cdot 0,5}{P(w)}=\frac{\frac{2}{6}}{P(w)}=\frac{1}{3\cdot P(w)}$$ where $$P(w)=P((w\land R)\lor(w\land B))=P(w\land R)+P(w\land B)=P(w\mid R)P(R) + P(w\mid B)P(B)=\frac{2}{3}\cdot 0,5 + \frac{3}{4}\cdot 0,5=\frac{2}{6} + \frac{3}{8}=\frac{17}{24}$$
Therefore, we get $$P(R\mid w)=\frac{1}{3\cdot \frac{17}{24}}=\frac{1}{\frac{17}{8}}=\frac{8}{17}$$
Is everything correct? (Wondering)

That looks all correct to me. (Happy)

Why do we use the Bayes' theorem to calculate $P(R\mid w)$ but not for $P(w\mid R)$ ? (Wondering)

Because we can tell what $P(w\mid R)$ is, knowing that we pick a random ball from R.
But we don't know what $P(R\mid w)$ is, other than using Bayes' theorem? (Thinking)

 

[mathmari]

That looks all correct to me. (Happy)
Because we can tell what $P(w\mid R)$ is, knowing that we pick a random ball from R.
But we don't know what $P(R\mid w)$ is, other than using Bayes' theorem? (Thinking)

Ah ok! Thank you so much! (Smile)