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I'm having a hard time with this question, but I did the best that I could. I would appreciate any help to correctly solve it.

Suppose that a coin is tossed three times and the side that lands up is noted. For instance, HHT indicates that the coin landed with a head up on the first two tosses and a tail up on the third.

        a)  List all the possible outcomes of the sample space.

        b)  Write each of the following events as a set and
            find its probability.

              i) exactly one toss results in a head

              ii) at least two tosses result in a head

             iii) the event that no head is obtained.

        c)  What is the probability that  exactly two
            tosses were heads if we know that there was
            a head on the first toss?

        d) What is the probability that exactly two heads
           were tosses if we know that at least one of the
           tosses was a head?

My answers:

A) 3 tosses and each toss has 2 possibilities, so: 2^3= 8 possibilities


B) (i) 1/8 = 12.5% (ii) 2/8 = 1/4 = 25% (iii) 1/8 = 12.5%

C) 1/8= 12.5%

D) 4/8 = 1/2 = 50% (The wording on this question is hard to decipher)

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Your answer to A is correct. In B, if you were to follow instructions and first make a list of all possible outcomes which result in one Head (winnow the list in your answer to A!) you will see why your answers to B(i) and B(ii) are wrong. – Dilip Sarwate Dec 15 '11 at 15:34
up vote 3 down vote accepted

Your answer to (a) is correct.

Hint: The set of events with "exactly one toss results in a head" is: $\{HTT, THT, TTH\}$. Given this what would be the answer to part (b) (i)?

You can do all the other questions if you carefully isolate the events so that they are consistent with the question text.

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A(i). HHT,HHT, HTH, HHT=4/8> 1/2 al head was first gotten

A(ii). HHH, HHT=2/8 >1/4 Dis ar d first two tossed result in a head

A(iii).. TTT=7/8 head was nt obtain in d above

A(iv)..HHH=1/8 Only dis contains two heads after first tossed

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