# Thread: Need help on probability with biasedness! *TOUGH QTN*

1. ## Need help on probability with biasedness! *TOUGH QTN*

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.

2. Originally Posted by zepher
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.

[snip]
a) Select one container at random then toss the two coins in it:

Pr(2 heads) = Pr( 2 heads | container 1) Pr(container 1) + Pr( 2 heads | container 2) Pr(container 2) = $(L_1)^2 \left( \frac{1}{2} \right) + (L_2)^2 \left( \frac{1}{2} \right) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 )$.

b) Toss one coin each from each container:

Pr(2 heads) = Pr(head from container 1) Pr(head from container 2) = $L_1 L_2$.

Consider the function $Q(L_1, L_2) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) - L_1 L_2 = \frac{1}{2} (L_1 - L_2)^2$.

You're interested in whether Q > 0 or Q < 0 .....

3. Originally Posted by mr fantastic

Consider the function $Q(L_1, L_2) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) - L_1 L_2 = \frac{1}{2} (L_1 - L_2)^2$.

You're interested in whether Q > 0 or Q < 0 .....
Actually, what's the $\frac{1}{2} (L_1 - L_2)^2$, I don't really get why I must equate to that. Shouldn't I equate $\frac{1}{2} ( (L_1)^2 + (L_2)^2 ) > L_1 L_2$ to prove that A is a better choice than B? But even so, how should I go about proving it?

4. Originally Posted by zepher
Actually, what's the $\frac{1}{2} (L_1 - L_2)^2$, I don't really get why I must equate to that. Shouldn't I equate $\frac{1}{2} ( (L_1)^2 + (L_2)^2 ) > L_1 L_2$ to prove that A is a better choice than B? But even so, how should I go about proving it?
If you looked more carefully at my reply you would note that:

1. The function Q is the difference between the two probabilities given by option a) and option b).

2. $\frac{1}{2} (L_1 - L_2)^2$ is the simplification of this function and therefore of the difference. In other words, the difference between the two probabilities given by option a) and option b) is $\frac{1}{2} (L_1 - L_2)^2$.

3. The difference is always greater than zero. Therefore the probability given by option a) is always greater than the probability given by option b) .......

5. Oh I got it now! But how did you simplify to get $\frac{1}{2} (L_1 - L_2)^2$ ? I can't seem to simplify to that. Sorry, I'm a bit dumb here =P

6. anybody can help? kinda urgent now =P

7. Originally Posted by zepher
Oh I got it now! But how did you simplify to get $\frac{1}{2} (L_1 - L_2)^2$ ? I can't seem to simplify to that. Sorry, I'm a bit dumb here =P
$Q(L_1, L_2) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) - L_1 L_2 = {\color{red}\frac{1}{2} ((L_1)^2 - 2L_1 L_2 + (L_2)^2)} = \frac{1}{2} (L_1 - L_2)^2$.

It's expected that you can factorise standard forms.