There are two coins labeled A and B Both coins are flipped b

There are two coins labeled A and B. Both coins are flipped, but one coin is chosen at random to be flipped first, and then the other coin is flipped. Coin A has probability 2/3 of landing heads. Coin B has probability 3/4 of landing heads. Define the following notation.

W = { 1 if coin A flipped first, 2 if coin B flipped first}

Y = { 1 if first coin flipped lands on heads, 0 is first coin flipped lands on tails }

Z = { 1 if second coin flipped lands on heads, 0 if second coin flipped lands on tails }

Since one coin is chosen at random to be flipped first, we have that

P = (W=1)=P(W=2)=1/2

We also have

P(X=1 | W = 1) = 2/3

and

P(X=1 | W=2) = 3/4

(a) Find P(X = 1). This is the probability that the first coin flipped lands heads.

(b) Find P(W = 1|X = 1). This is the probability that the first coin flipped is coin A, conditional on the coin landing heads.

(c) Find P(Z = 1|X = 1). This is the probability that the second coin flipped lands heads, conditional on the first coin flipped landing heads.

(d) Show (mathematically) that X and Z are not independent.

Solution

(a) P(X = 1) = P(X=1 | W = 1)*P(W = 1) + P(X=1 | W=2)*P(W = 2)

= (2/3)*(1/2) + (3/4)*(1/2)

= 17/24

= 0.708

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(b) P(W = 1|X = 1) = P(X=1 | W = 1) / P(X = 1) {Accoriding to Bayes theorem}

= (2/3) / (17/24)

= 16/17

= 0.9412

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P(Z = 1|X = 1) = P(X = 1|Z = 1)/ P(X = 1)

Now, we find P(X = 1|Z = 1):

P(X = 1|Z = 1) = P(X = 1 and Z = 1)/P(Z = 1)

= (1/2)*{2/3 + 3/4} + (1/2)*{3/4 + 2/3}

= 1/2

P(Z = 1|X = 1) = (1/2) / (17/24)

= 12/17

= 0.7059

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Show (mathematically) that X and Z are not independent.

Let us take the example P(X = 1 and Z = 1)

If X and Z are independent, then P(X = 1 and Z = 1) = P(X = 1) * P(Z = 1)

P(X = 1 and Z = 1) = 1/2

P(X = 1) = 17/24

P(Z = 1) = (2/3)*(1/2) + (3/4)*(1/2) = 17/24

As it can be seen that P(X = 1 and Z = 1) is not equal to P(X = 1) * P(Z = 1). Therefore, X and Z are independent.