Suppose now I have two BIASED coins in two containers (4 coins in total).
Container 1's coins have a probability of L1 of getting heads and Container 2's coins have a probability of L2 of getting heads. L1 is not equal to L2.
Which will give a higher chance of getting two heads if I:
a) Select one container at random then toss the two coins in it
b) Toss one coin each from each container
Please guide me and give me hints. I want to come up with the solution myself!
For a start, this is about conditional probability, right?
Also, I think option A will give a higher chance, but I'm not sure if I'm correct.
IF I were to use imaginary figures taking L1 as 0.9 and L2 as 0.4, by doing:
a) Pr1 = (0.9 * 0.9) * 0.5 = 0.405 | Pr2 = (0.4 * 0.4) * 0.5 = 0.08
b) Pr = 0.9 * 0.4 = 0.36 (< Pr1)
Option A is better in this case
If L1 is 0.5 and L2 is 0.3,
a) Pr1 = (0.5 * 0.5) * 0.5 = 0.125 | Pr2 = (0.3 * 0.3) * 0.5 = 0.045
b) Pr = 0.5 * 0.3 = 0.15 (> Pr1)
Option B is better in this case
After here, I'm stuck and unsure how to proceed next to justify my choice. Please help! ^^ Thanks.