lheiannie07 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

How can i start solving this? Can any experts help me to formulate the solution to this? Thanks

OA D

Say the probability of getting a head is p(h) and the probability of getting a tail is p(t), thus,

Probability of getting a tail on the first flip = p(t) = 1/2;
Probability of getting a tail on the second flip = p(h)*p(t) = 1/2*1/2 = 1/4;
Probability of getting a tail on the third flip = p(h)*p(h)*p(t) = 1/2*1/2*1/2 = 1/8

Probability of getting a tail in the first three flips = 1/2 + 1/4 + 1/8 = 7/8.

We are interested in the probability of getting a tail on the THIRD flip (1/8) given that the probability of getting a tail in the first three flips is 7/8.

Thus, the required probability = (1/8) / (7/8) = 1/7.

The correct answer: D

Hope this helps!

-Jay

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BTGmoderatorDC wrote: ↑

Sun Oct 01, 2017 2:05 pm

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

How can i start solving this? Can any experts help me to formulate the solution to this? Thanks

OA D

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