Problem 2 (25 points) We toss three fair coins simultaneously and independently.

Question

Problem 2 (25 points) We toss three fair coins simultaneously and independently. If the outcomes of all coin tosses are the same, we win; otherwise, we lose. Let A be the event that the first coin and second coin come up heads, B be the event that the second coin and third coin come up heads, and C be the event that we win. Determine whether the below events are true or false. Events A and B are not independent. Events A and C are independent. Events A and B are conditionally independent given C. Events A and C are conditionally independent given B . The probability of winning 1

Explanation / Answer

Ans Let the event of head appearing in ith coin be Hi and that tail appearing be Ti .

Therefore P(A) = P(H1 and H2) and since outcome of each coin is independent

= P(H1) * P(H2) = 0.5 * 0.5 = 0.25

Similarly P(B) = P(H2 and H3)

= P(H2 ) * P( H3) = 0.5 *0.5 = 0.25

A and B are independent if P(A and B) = P(A) * P(B)

P(A and B) = P((H1 and H2) and (H2 and H3 ) )

= P(H1 and H2 and H3 )

= 0.5 * 0.5 * 0.5 = 0.125

But P(A) * P(B) = 0.25 * 0.25 = 0.0625

Therefore Events A and B are not independent, so first event is true.

C is the event (H1 and H2 and H3 ) or (T1 and T2 and T3 )

P(C) = P(H1 and H2 and H3 ) + P(T1 and T2 and T3 )

= 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25

A and C is the event that first two coins have head and we win meaning all of the coins have heads

Therefore P(A and C) = P(H1 and H2 and H3 ) = 0.125

But P(A) * P(C) = 0.25 * 0.25 = 0.0625

Therefore A and C are not independent events so second event is false.

Events A and B are conditionally independent given C if

P(A and B | C) = P(A|C) * P(B|C)

From the formula for Conditional Probability

P(A and B | C) = P(A and B and C) / P(C)

A and B and C basically imply that all the coins shows head

Therefore  P(A and B and C) = P(H1 and H2 and H3 ) = 0.125

And P(C) = 0.25

P(A and B | C) = P(A and B and C) / P(C) = 0.5

And P(A|C) is the probability that we have win and A has occured implies we have won using all heads probability of which is 0.5 since we had only two ways of to win each with equal probability.

Similarly P(B|C) = 0.5

So P(A|C) * P(B|C) = 0.25 which is not equal to P(A and B | C) so

Events A and B are not conditionally independent given C.

Similarly events A and C are conditionally independent given B if

P(A and C | B) = P(A|B) * P(C|B)

From the formula for Conditional Probability

P(A and C | B) = P(A and B and C) / P(B)

A and B and C basically imply that all the coins shows head

Therefore  P(A and B and C) = P(H1 and H2 and H3 ) = 0.125

And P(B) = 0.25

P(A and C |B) = P(A and B and C) / P(B) = 0.5

And P(A|B) = P(A and B) / P(B) = 0.125 / 0.25 = 0.5

And P(C|B) is the probability that we win given that second and third coin has head which means first coin will have head the probability of which is 0.5

So P(A|B) * P(C|B) = 0.25 which is not equal to P(A and C | B) so

Events A and C are not conditionally independent given B.

The probability of winning is 0.25 = 2/8 so the last given event is false.

Please upvote the answer if found helpful or do comment for any further help.

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