Harvey flipped a fair coin until he got tails, at which point he stopp

Question

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.  \frac{1}{2} 
 
B.  \frac{1}{3} 
 
C.  \frac{1}{4} 
 
D.  \frac{1}{7} 
 
D.  \frac{1}{8}

KarishmaB · Accepted Answer

P (T on first flip) = 1/2
P (T on second flip but not on first - HT) = (1/2)*(1/2) = 1/4
P (T on third flip but not on first two - HHT) = (1/2)*(1/2)*(1/2) = 1/8
 P(T on exactly one of the first three flips and stop) = 1/2 + 1/4 + 1/8 = 7/8

According to conditional probability:

P (A given B) = P(A)/P(B)

P ("T on third flip but not on first two" given "T on exactly one of the first three flips and stop") = (1/8) / (7/8) = 1/7

Answer (D)

ScottTargetTestPrep · Answer

Harvey stopped after no more than three flips 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

arunavamunshi1988 · Answer

Hi,

I did not understand the application of this conditional probability formula. What I understand from the problem,

Probability = No of Desired Cases/ Total no of cases.

Now number of desired cases of getting Tail on the third flip only = 1
Total no of cases = 2*2*2 = 8

So probability should be = 1/8...

Please let me know what I am missing here? If possible, please explain the ans without the application of this formula because I prefer to solve the problems of probability without using any formula.

AnthonyRitz · Answer

Close, but since HHH is impossible there are only 7 total possible cases. Thus 1/7 instead of 1/8.

kkrrsshh · Answer

AnthonyRitz  Could you please tell why HHH is impossible?

AnthonyRitz · Answer

The question itself provides that Harvey stops flipping after no more than three flips. Since he stops only on a Tails, he must have gotten a Tails within those flips.

ShashankDave · Answer

Hi  VeritasPrepKarishma

Conditional Probability is exactly how I did it. However, I want to know how to do this one by using purely permutations and combinations. Would be great help.

DwightSchrute · Answer

wouldn't the probability of getting exactly one tales out of the 3 flips be 3/8?
THH
HTH
HHT

and then we can use the formula P(A)/P(B) = 1/8/3/8 = 1/3

Please let me know what I am doing wrong

gh2021 · Answer

He stops after he gets tails.

Example: THH is not a case since if he gets Tails first time, he stops. So the outcomes are -
T
HT
and HHT

Also, corresponding probabilities are 1/2, 1/4 and 1/8

total = 7/8

Out of these 7/8 outcomes, we have to find the probability of 1/8 (3rd outcome), so 1/8 divided by 7/8 = 1/7

Hahahehe · Answer

Why is everyone using the incorrect formula for conditional probability, including  VeritasKarishma  ?

Can someone please comment if my answer and understanding are correct?

The correct formula for conditional probability is:
P ( A | B ) = P (A and B)/ P(B) , where P(A | B ) means probability of event A, given B has happened, P(A and B) means probability of event A and B both happening

Now, here, event A = events with Tails only at the 3rd place. B = events where Harvey got tails within the first three flips.

P(A) = 1/8 as out of the 8 possible combinations, only 1 possible case where tails is the at 3rd place and nowhere else. (HHT)
P(B) = 7/8 as out of the 8 possible combinations, 7 have Tails at one place or the other. (all except HHH)
P(A and B) = 1/8 as again, only 1 outcome satisfies both A and B. (HHT)

Now P(A | B) = P(A and B) / P(B) = (1/8)/(7/8) = 1/7

Hahahehe · Answer

VeritasKarishma ,  AnthonyRitz ,

If we are told that harvey stopped at no more than 3 tosses, doesn't it leave us with just the following possibilities :
T
HT
HHT

and given this, only 1 possibility has T at the third place, so shouldn't the answer be just 1/3 ? Can someone explain where exactly is my thinking wrong

AnthonyRitz · Answer

It is true that these are the only outcomes that are really possible, in a certain manner of approaching it. The problem is that T (1/2) is twice as likely as HT (1/4), which, in turn, is twice as likely as HHT (1/8).

That’s why you can’t just add them and get 1/3.

Posted from my mobile device

That’s why you can’t just add them and get 1/3.

Posted from my mobile device

KarishmaB · Answer

Since A is a more restricted event under event B, 
P(A given B) = P(A)/P(B)

In other words, think in terms of sets - A is a subset of B.

Usernamevisible · Answer

Doubt with solution to it

Doubt 
There are only three possible scenarios if Harvey stops within three flips: T, HT, and HHT. Since we want the cases where he did not get tails on the third flip, wouldn't the favorable cases be T and HT, giving a probability of 2/3?

Also, I remember solving other probability questions by first restricting the sample space to all valid cases under the given condition, treating that as the new universe, and then counting the favorable cases within that universe. Why does that approach not work here?

Solution 
Yes, but that works only when the cases in your population are equally likely.

You are remembering the rule:

P(E) = (Number of favorable cases) / (Total number of possible cases)

This formula is valid only if every case in the denominator has the same probability.

---

Here, the three stopping outcomes are:

T with probability 1/2
HT with probability 1/4
HHT with probability 1/8

These are not equally likely, so you cannot say:

P(HHT) = 1/3

---

When counting cases DOES work:

Suppose the question were:

A fair coin is flipped exactly twice. What is the probability of getting exactly one tail?

The sample space is:

HH
HT
TH
TT

Each outcome has probability 1/4, so counting works:

P(exactly one tail) = 2/4 = 1/2

---

Another example with conditioning:

If we know that at least one head occurred in two flips, the reduced sample space becomes:

HH
HT
TH

These three outcomes are still equally likely because each originally had probability 1/4, so we can count within this new universe.

For example:

P(both heads | at least one head) = 1/3

---

The key test is:

Whenever you create a new "universe" after conditioning, ask:

Are all the remaining cases equally likely?

If yes, simple counting is valid.

If no, use probabilities (weights), as in this coin-stopping problem.

Here, T, HT, and HHT have probabilities 1/2, 1/4, and 1/8, respectively, so:

P(HHT | stop in <= 3 flips) = (1/8) / (1/2 + 1/4 + 1/8) = 1/7.

Harvey flipped a fair coin until he got tails, at which point he stopp

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Harvey flipped a fair coin until he got tails, at which point he stopp

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