Tricky probability question

  1. A coin is tossed 3 times. at least 1 head is obtained. Determine each probability...?

    1) exactly 1 head is obtained
    2) exactly 2 heads are obtained
    3) exactly 3 heads are obtained.
    ------------------------------------------------------------

    for 1) brute force indicates that there 7 possible combinations:

    HHH, HHT, HTH, HTT, THH, THT, TTH (because at least one heads is obtained).

    Out of these we see that there are 3 occasions where there is exactly one heads.
    If we let S be the sample space of all possable outcomes, and let E be the event of having exactly one heads, then a basic rule of probability indicates that the probability P(E)=|E|/|S|=3/7.

    However, using another method:
    fix one coin throw's result as heads. Then the probability of the other two throws being tails is (1/2)(1/2)=1/4.
    Since we can fix heads 3 times, this indicates that the probability is 3(1/4)=3/4.

    Can anyone spot where the flaw is in either of these attempts at a solution?
    thanks
    (this isn't a hw question btw)
     
  2. jcsd
  3. EnumaElish

    EnumaElish 2,481
    Science Advisor
    Homework Helper

    3/7 is a conditional probability given at least one head. For the unconditional probability you need to count TTT among the outcomes.

    3/4 is the conditional probability of two tails given (one) head. "Head" has probability 1/2, so the unconditional probability is 3/4 x 1/2 = 3/8.
     
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