Harvey flipped a fair coin until he got tails, at which point he stopped. If Harvey stopped after no more than three flips, what is the probability that he didn't get tails until his third flip?
A. 1/2
B. 1/3
C. 1/4
D. 1/7
E. 1/8
Here the OA is D
My Approach: The question says he didn't get tails until his third flip. In this case the scenario is HHT.
Probability of coins being flipped three times is 1/8.
Since we have just one set 1*(1/8) = 1/8.
What is wrong in the above approach?
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Kaustubhk wrote:Harvey flipped a fair coin until he got tails, at which point he stopped. If Harvey stopped after no more than three flips, what is the probability that he didn't get tails until his third flip?
A. 1/2
B. 1/3
C. 1/4
D. 1/7
E. 1/8
Here the OA is D
My Approach: The question says he didn't get tails until his third flip. In this case the scenario is HHT.
Probability of coins being flipped three times is 1/8.
Since we have just one set 1*(1/8) = 1/8.
What is wrong in the above approach?
The way to look at this question is in terms of conditional probability. What is the probability of T happening in 3rd round given that the person would have stopped in any of the 3 rounds due to a T.
A: Tails happen only in 3rd round => HHT = 1/8
B: Tails happens in either round 1 or round 2 or round 3 ==> T, HT, HHT ==> 1/2 + 1/4 + 1/8 = 7/8.
Note: (A n B) = HHT => P(A n B) = 1/8
Conditional probability:
P(A/B) = P(AnB) / P(B) = 1/8 / 7/8 => 1/7
Hence D.
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Hi Kaustubhk,
This question is NOT a typical probability question, so you have to note the specific language that is used and then do the appropriate work/math. As an aside, it's highly unlikely that you will be asked about this on the Official GMAT.
We're told that Harvey flips a coin UNTIL he gets 'tails' (at which point he stops) and that he flipped no more than 3 times. Here are the possible outcomes based on THAT description (and the probability of each individual outcome occurring).
T = 1/2
HT = (1/2)(1/2) = 1/4
HHT = (1/2)(1/2)(1/2) = 1/8
1/2 + 1/4 + 1/8 = 7/8
In simple terms, you can take this result to mean "with up to 3 tosses, there are 8 outcomes and 7 of them 'end in tails'." Of those 7 outcomes that 'end in tails', only 1 of them (re: HHT) matches the result that we're looking for. Thus, the probability of this specific event  under the specific condition that he stops flipping once he hits 'tails'  is 1/7.
Final Answer: D
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This question is NOT a typical probability question, so you have to note the specific language that is used and then do the appropriate work/math. As an aside, it's highly unlikely that you will be asked about this on the Official GMAT.
We're told that Harvey flips a coin UNTIL he gets 'tails' (at which point he stops) and that he flipped no more than 3 times. Here are the possible outcomes based on THAT description (and the probability of each individual outcome occurring).
T = 1/2
HT = (1/2)(1/2) = 1/4
HHT = (1/2)(1/2)(1/2) = 1/8
1/2 + 1/4 + 1/8 = 7/8
In simple terms, you can take this result to mean "with up to 3 tosses, there are 8 outcomes and 7 of them 'end in tails'." Of those 7 outcomes that 'end in tails', only 1 of them (re: HHT) matches the result that we're looking for. Thus, the probability of this specific event  under the specific condition that he stops flipping once he hits 'tails'  is 1/7.
Final Answer: D
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You're on the right track, but you're ignoring one key condition: Harvey might've stopped earlier.Kaustubhk wrote: My Approach: The question says he didn't get tails until his third flip. In this case the scenario is HHT.
Probability of coins being flipped three times is 1/8.
Since we have just one set 1*(1/8) = 1/8.
What is wrong in the above approach?
With that in mind, we think of probability as
Target Outcome / All Possible Outcomes
Our target = HHT, and all possible outcomes = T, HT, and HHT. From here, we have 1/8 / (1/2 + 1/4 + 1/8), or 1/8 / 7/8, or 1/7.
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Harvey stopped if he had any of the following outcomes:Kaustubhk wrote: ↑Sun May 21, 2017 1:10 amHarvey flipped a fair coin until he got tails, at which point he stopped. If Harvey stopped after no more than three flips, what is the probability that he didn't get tails until his third flip?
A. 1/2
B. 1/3
C. 1/4
D. 1/7
E. 1/8
Here the OA is D
My Approach: The question says he didn't get tails until his third flip. In this case the scenario is HHT.
Probability of coins being flipped three times is 1/8.
Since we have just one set 1*(1/8) = 1/8.
What is wrong in the above approach?
T, HT, HHT
The probability of the first outcome is P(T) = 1/2. The probability of the second outcome is P(HT) = 1/2 x 1/2 = 1/4. The probability of the third outcome is P(HHT) = 1/2 x 1/2 x 1/2 = 1/8.
Since only the third outcome is the one of interest, the probability that the third outcome occurs, given that any of the three outcomes can happen, is:
(1/8)/(1/2 + 1/4 + 1/8) = (1/8)/(7/8) = 1/7
Answer: D
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