Conditional Probability and law of total probability.

[luke95]

This is a fun question i found on the internet, a bit harder than my course and I've spent hours on it but can't find a solution, i was hoping someone could help me.
Here's the situation.
You are in jail and have been sentenced to death tomorrow, however there's a way out.
you're given 12 red balls and 12 blue balls with 2 urns, you have to fully distribute the balls between these two urns. the executioner will then select an urn and draw a ball
if it is blue you live and if it is red you die.
suppose you place k balls into the first urn and 24-k into the second
suppose you put i blue balls into the first urn and 12-i into the second
Let A be the event that you live.
By conditioning on the urn that the executioner chooses, and using the law of total probability show that:

P(A) = (12i - ki + 6k) / k(24-k)

please could you explain the steps used so i can try and understand the solution and method.
Thank you in advance
Luke

 

[chisigma]

This is a fun question i found on the internet, a bit harder than my course and I've spent hours on it but can't find a solution, i was hoping someone could help me.
Here's the situation.
You are in jail and have been sentenced to death tomorrow, however there's a way out.
you're given 12 red balls and 12 blue balls with 2 urns, you have to fully distribute the balls between these two urns. the executioner will then select an urn and draw a ball
if it is blue you live and if it is red you die.
suppose you place k balls into the first urn and 24-k into the second
suppose you put i blue balls into the first urn and 12-i into the second
Let A be the event that you live.
By conditioning on the urn that the executioner chooses, and using the law of total probability show that:

P(A) = (12i - ki + 6k) / k(24-k)

please could you explain the steps used so i can try and understand the solution and method.
Thank you in advance
Luke

Wellcome on MHB luke 95!... You have i blue balls over k balls in the first urn and 12 - i blue balls over 24 - k balls in the second urn. If You suppose that the probability that executioner chooses first or second urn is the same [$\frac{1}{2}$], being the favourable probability $\frac{i}{k}$ for the first urn and $\frac{12 - i}{24 - k}$ for the second urn, is... $\displaystyle PA = \frac{1}{2}\ \frac{i}{k} + \frac{1}{2}\ \frac{12 - i}{24 - k} = \frac{12\ i - k\ i + 6\ k}{k\ (24 - k)}\ (1)$ Kind regards $\chi$ $\sigma$

 

[luke95]

Wellcome on MHB luke 95!... You have i blue balls over k balls in the first urn and 12 - i balls over 24 - k balls in the second urn. If You suppose that the probability that executioner chooses first or second urn is the same [$\frac{1}{2}$], being the favourable probability $\frac{i}{k}$ for the first urn and $\frac{12 - i}{24 - k}$ for the second urn, is... $\displaystyle PA = \frac{1}{2}\ \frac{i}{k} + \frac{1}{2}\ \frac{12 - i}{24 - k} = \frac{12\ i - k\ i + 6\ k}{k\ (24 - k)}\ (1)$ Kind regards $\chi$ $\sigma$

thank you very much :)

 

[chisigma]

Wellcome on MHB luke 95!... You have i blue balls over k balls in the first urn and 12 - i blue balls over 24 - k balls in the second urn. If You suppose that the probability that executioner chooses first or second urn is the same [$\frac{1}{2}$], being the favourable probability $\frac{i}{k}$ for the first urn and $\frac{12 - i}{24 - k}$ for the second urn, is... $\displaystyle PA = \frac{1}{2}\ \frac{i}{k} + \frac{1}{2}\ \frac{12 - i}{24 - k} = \frac{12\ i - k\ i + 6\ k}{k\ (24 - k)}\ (1)$ Kind regards $\chi$ $\sigma$

The question proposed by luke95 is 'veryvery interesting' because it leads to a paradoxical conclusion [not unusual in probability ...]. Anybody at first can imagine that at least the chance to survive can be one half but it isn't true. If You look at the (1) in my post You realize that choosing i=k in any case the probability of survive is greater. With a simple further analysis You discover that PA has its maximum for i= k = 1 and in this case is $PA = \frac{17}{23} \sim .739$... incredible!...

Kind regards

$\chi$ $\sigma$

 

[CellCoree]

I understand the importance of probability and how it can be used in various situations. In this scenario, we are dealing with conditional probability, which is the probability of an event occurring given that another event has already occurred. In this case, the event of living (A) is dependent on the urn that the executioner chooses and the color of the ball drawn.

To find the probability of living (A), we can use the law of total probability, which states that the total probability of an event is equal to the sum of the probabilities of all possible outcomes. In this case, we have two possible outcomes – the executioner choosing the first urn or the second urn.

Let's start by finding the probability of living (A) given that the executioner chooses the first urn (U1). We can represent this as P(A|U1). Since there are a total of 12 blue balls in the first urn, the probability of drawing a blue ball and living is i/12. Similarly, the probability of drawing a red ball and dying is (12-i)/12. Therefore, the total probability of living given that U1 is chosen is:

P(A|U1) = i/12

Now, let's find the probability of living given that the executioner chooses the second urn (U2). In this case, there are a total of 12-k blue balls in the second urn, so the probability of drawing a blue ball and living is (12-k)/12. The probability of drawing a red ball and dying is k/12. Therefore, the total probability of living given that U2 is chosen is:

P(A|U2) = (12-k)/12

Next, we can use the law of total probability to find the overall probability of living (P(A)). This is given by:

P(A) = P(A|U1)*P(U1) + P(A|U2)*P(U2)

= (i/12)*(1/2) + ((12-k)/12)*(1/2) [since there are equal chances of choosing either urn]

= (i + 12-k)/24

= (12i - ki + 6k)/24

= (12i - ki + 6k)/(k + (24-k)) [since there are k balls in the first urn and 24-k balls in the second urn]

= (12i - ki +