SOLUTION: three biased coins C1,C2,C3l ie on the table .Their probabilities of falling heads are 1/3,2/3 and1 respectiely . A coin is picked at random, tossed twice ,and obser

Question 1197263: three biased coins C1,C2,C3l ie on the table .Their probabilities of falling heads are 1/3,2/3 and1 respectiely . A coin is picked at random, tossed twice ,and observed to fall heads both times .Calculate the probability that it is C1.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52835) (Show Source): You can put this solution on YOUR website!
.
three biased coins C1, C2, C3 lie on the table.
Their probabilities of falling heads are 1/3, 2/3 and 1, respectively.
A coin is picked at random, tossed twice, and observed to fall heads both times.
Calculate the probability that it is C1.
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They want you calculate the conditional probability that the randomly picked coin is C1 given that tossed twice it produces (Head,Head). P(tossing one of the three coins twice gives HH) = = = = = . P(choosing C1 and (tossing C1 twice gives HH)) = = . Therefore, the ANSWER is P = P(choosing C1 and (tossing C1 twice gives HH)) / P(tossing one of the three coins twice gives HH) = = = .

Solved, with full explanations.

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

The tutor @ikleyn has a great solution.
My approach is the same idea, just written slightly different.
There's also an alternative visual way to see how this all works.

H = heads
T = tails

Let's define these four events
A = coin is C1
B = coin is C2
C = coin is C3
D = result is HH

We're asked to find P(A given D)

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First let's take a slight detour

P(A) = P(B) = P(C) = 1/3
assuming each coin is equally likely to be selected
note that
P(A)+P(B)+P(C) = 1

Then calculate the conditional probability of getting HH for each coin
P(D given A) = P(H given A)*P(H given A) = (1/3)*(1/3) = 1/9
P(D given B) = P(H given B)*P(H given B) = (2/3)*(2/3) = 4/9
P(D given C) = P(H given C)*P(H given C) = 1*1 = 1

Those 6 items will then help us determine P(D)

Use the Law of Total Probability to find the following
P(D) = P(D and A)+P(D and B)+P(D and C)
P(D) = P(D given A)*P(A)+P(D given B)*P(B)+P(D given B)*P(B)
P(D) = (1/9)*(1/3)+(4/9)*(1/3)+(1)*(1/3)
P(D) = 1/27 + 4/27 + 1/3
P(D) = 1/27 + 4/27 + 9/27
P(D) = (1+4+9)/27
P(D) = 14/27
Thing to notice: In the diagram below, exactly 14 rectangles are shaded yellow out of 9*3 = 27 total.
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Now we can calculate what we're after.
P(A given D) = P(A and D)/P(D)
P(A given D) = P(D given A)*P(A)/P(D)
P(A given D) = (1/9)*(1/3)/(14/27)
P(A given D) = (1/9)*(1/3)*(27/14)
P(A given D) = (1*1*27)/(9*3*14)
P(A given D) = (1*1*27)/(27*14)
P(A given D) = 1/14
which is the final answer.

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Here's a visual way to think about it

C1,C2,C3 are equally likely. They represent the three columns.
The numbers inside the yellow boxes help us keep track of each column's values.
Each yellow rectangle represents the outcome HH, while a white nonshaded rectangle is some other outcome (HT, TH or TT).

Column C1 has 1 yellow rectangle shaded out of 9. This is to represent this coin's probability of getting two heads in a row.

Column C2 has 4 yellow rectangles shaded out of 9, showing that 4/9 is the probability of getting HH for coin C2.

Column C3 has all 9 rectangles shaded since we're guaranteed to get two heads for this coin.

There are 1+4+9 = 14 rectangles shaded yellow.

Now imagine we know 100% that we got two heads in a row. That must mean we know we landed somewhere on a yellow rectangle, or we know for certain we have pulled a yellow rectangle out of the bag.

C1 only chips in one such shaded rectangle, so that's another way to see how the answer is 1/14

If we wanted to know P(B given D), then the answer would be 4/14 since C2 chips in 4 yellow rectangles out of the 14 total.

Lastly P(C given D) = 9/14 for similar reasoning.

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Answer: 1/14

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