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Please could someone verify if my proposed solutions stack up correctly for the following questions?

Consider the tossing of two fair coins:

a) Compute the probability of at least one head and a match or both? Sol: the sample set of outcomes will be {(HH),(HT),(TH),(TT)}.

Pr(atLeastOneHead) = 3/4 = 0.75

Pr(match) = 2/4 = 0.5

Pr(both) = 3/4 + 1/2 - 1/4 = 1

b) What is the conditional probability of obtaining two heads when flipping a coin twice given that at least one head was obtained?

Using the sample space => {(HH),(HT),(TH),(TT)}. The pr(atLeastOneHeadObtained) = 3 out of the four outcomes = 3/4

Pr((twoHeadsObtainted) n (atLeastOneHeadObtained)) = 1/4

The conditional probability = .25 / .75

c) Are the events that at least one head shows and a match statistically independent?

Two events are independent if P(AnB) = P(A).P(B).

P(AnB) = 1/4

P(A) * P(B) = 1/2 * 1/2 = 1/4

Therefore, the events are statistically independent

Thanks in advance

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up vote 2 down vote accepted

a) Wrong answer. In calculating $P(A \cup B)$, you need to use the principle of Inclusion and Exclusion which states that

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

b) Correct answer. Correct approach.

c) Wrong answer. Correct approach. Calculate $P( A \cap B)$ again, it is not 0. For example, it is possible to have at least 1 head AND get a match of the coins? (This might tie back to part a.)

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Thanks for the feedback. For a), I should have Pr(atLeastOneHead) = 2/4 = 0.75 = {(HH),(HT),(TH)}. Pr(match) = 2/4 = 0.5 = {(HH),(TT)}. So Pr(both) = 3/4 + 1/2 - 1/4. – bosra Jan 8 '13 at 17:21
@bosra Note that Pr(at least one head) = 3/4 (you keep on saying 2/4 or 0.5). Why don't you go and edit your question, so that anyone else who looks at it will now know the correct answer? – Calvin Lin Jan 8 '13 at 18:13

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