An unfair coin with pr[h] = 0.2 is flipped. if the flip results in a head, a marble is selected at random from a urn containing six red and four blue marbles. otherwise, a marble is selected from a different urn containing three red and five blue marbles. if the selected marble selected is red, what is the probability that the flip resulted in a head?

Normally when dealing with coins the probability of getting heads or tails is 0.5 each. However in this case since its an unfair coin, the probability of getting heads is 0.2. 
H - head 
T - tails
R - red marble
pr (H) = 0.2
urn
6 red and 4 blue
pr (T)   = 0.8
urn
3 red and 5 blue

when heads is obtained 
red - 6/10 -0.6
blue - 4/10 - 0.4
therefore when multiplying with 0.2 probability of getting heads
pr (R ∩ H) red - 0.6*0.2 = 0.12

when tails is obtained 
red - 3/8 - 0.375
blue - 5/8 - 0.625
when multiplying with 0.8 probability of getting tails
pr (R ∩ T) red - 0.375 * 0.8 = 0.3

using bayes rule the answer can be found out, 
the following equation is used;
pr (H | R) = pr (R ∩ H) / {pr (R ∩ H) + pr (R ∩ T)}
               = 0.12 / (0.12 + 0.3)
               = 0.12 / 0.42
               = 0.286
the final answer is 0.286

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .

Hey dear, hoping i'll be