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

  • $\begingroup$ For those confused reading the selected answer, the question has been edited to show the correct approach. The answer is for the previous version of the question which showed the incorrect solutions for a) and c) $\endgroup$ – Kelly Bang Jul 8 at 18:15

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.)

  • $\begingroup$ 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. $\endgroup$ – bosra Jan 8 '13 at 17:21
  • $\begingroup$ @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? $\endgroup$ – Calvin Lin Jan 8 '13 at 18:13

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