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Tricky probability question 
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#1
May2510, 03:34 AM

P: 19

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
May2510, 01:45 PM

Sci Advisor
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P: 2,481

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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