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Probability-Coins

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Probability-Coins

by Kaustubhk » Sun May 21, 2017 1:10 am
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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by [email protected] » Sun May 21, 2017 5:04 am
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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by [email protected] » Sun May 21, 2017 9:58 am
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.

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by [email protected] » Wed May 24, 2017 5:28 pm
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?
You're on the right track, but you're ignoring one key condition: Harvey might've stopped earlier.

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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Re: Probability-Coins

by [email protected] » Sun Mar 15, 2020 7:59 am
Kaustubhk wrote:
Sun May 21, 2017 1:10 am
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?
Harvey stopped if he had any of the following outcomes:

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