MATH SOLVE

9 months ago

Q:
# 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?

Accepted Solution

A:

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

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