Problem Solving

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

GMAT assassins aren't born, they're made,
Rich

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

Answer: D