MHB 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

$\chi$ $\sigma$