Question
2.12
Exercises 45 Show that the above formula for the probability of the union of two events can be generalized to three events as follows: Pr(A U B C) = Pr(A) + Pr(B) + Pr(C) 2.9 2.10 Prove Theorem 2.3 which states that if AC B then Pr(A) S Pr(B) 2.11 Formally prove the union bound which states that for any events A, A, AM (not necessarily mutually exclusive), 2.12 An experiment consists of tossing a coin twice and observing the sequence of coir tosses. The sample space consists of four outcomes 1 (H,H), 2-(H, T). 3 = (T, H) , and 4-(T, T) . Suppose the coin is not evenly weighted such that we expect a heads to occur more often than tails and as a result, we assign the following probabilities to each of the four outcomes: , Pr( H, T) =1, Pr(T, H) = 1, Pr(T, T)- Pr( H, H) Does this probability assignment satisfy the three axioms of probability? Given this probability assignment, what is Pr(first toss is heads)? Given this probability assignment, what is Pr(second toss is heads) ? 4 4 (a) (b) (c) 2.13 Repeat Exercise 2.12 if the probability assignment is changed to: = 2 , Pr(H, T)-15, Pr(TH)-15, 64 Pr(T, T) = 2 64 Pr(H, H) 2.14 Consider the experiment of tossing a six-sided die as described in Example 2.2. Suppose the die is loaded and as such we assign the following probabilities to each of the six outcomes 21 21 Is this assignment consistent with the three axioms of probability
Explanation / Answer
Question 2.12
a) Yes it satisfies all the axioms of probability because all the probability values are greater than equal to 0 and less than equal to 1. Also the sum of all 4 probabilities is equal to 1. Also we are covering all the possible 4 outcomes on a toss of coin two times.
b) P( first toss is head ) is computed as:
P({ HH, HT } ) = (3/8) + (1/4) = 0.625
Therefore 0.625 is the required probability here.
c) P( second toss is a head ) is computed as:
P({ TH, TH } ) = (1/4) + (1/8) = 0.375
Therefore 0.375 is the required probability here.
