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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. \(\frac{1}{8}\)


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)

For more on conditional probability, check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... onditions/

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.
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VeritasPrepKarishma
ziyuen
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. \(\frac{1}{8}\)


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)

For more on conditional probability, check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... onditions/

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.

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


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

AnthonyRitz Could you please tell why HHH is impossible?
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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.
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VeritasPrepKarishma
ziyuen
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. \(\frac{1}{8}\)


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)

For more on conditional probability, check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... onditions/

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.
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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
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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

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
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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
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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
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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
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Hahahehe
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

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.
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