GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Aug 2019, 13:52 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # M21-11

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 57025
M21-11  [#permalink]

### Show Tags

2
10 00:00

Difficulty:   45% (medium)

Question Stats: 58% (01:13) correct 42% (01:24) wrong based on 126 sessions

### HideShow timer Statistics

If a coin is tossed until the sequence $$HTH$$ appears ($$H$$ denotes the fall of heads, $$T$$ denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. $$\frac{1}{16}$$
B. $$\frac{3}{32}$$
C. $$\frac{1}{8}$$
D. $$\frac{3}{16}$$
E. $$\frac{1}{4}$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
M21-11  [#permalink]

### Show Tags

Official Solution:

If a coin is tossed until the sequence $$HTH$$ appears ($$H$$ denotes the fall of heads, $$T$$ denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. $$\frac{1}{16}$$
B. $$\frac{3}{32}$$
C. $$\frac{1}{8}$$
D. $$\frac{3}{16}$$
E. $$\frac{1}{4}$$

We look for either $$HHTH$$ or $$THTH$$ sequence. The probability of either is $$(\frac{1}{2})^4 = \frac{1}{16}$$. The combined probability is $$\frac{2}{16} = \frac{1}{8}$$.

Answer: C
_________________
Director  Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 732
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)
Re: M21-11  [#permalink]

### Show Tags

I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4
_________________
Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: M21-11  [#permalink]

### Show Tags

PiyushK wrote:
I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4

The game will end after the fourth throw means that the fourth throw should be the last in the game.
_________________
Director  Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 732
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)
Re: M21-11  [#permalink]

### Show Tags

2
Bunuel wrote:
PiyushK wrote:
I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4

The game will end after the fourth throw means that the fourth throw should be the last in the game.

Then sentence should be:
what is the probability that the game will end with 4th throw?
what is the probability that the game will end at fourth throw?

After 4th throw is ambiguous; I discussed this question with other people and they also inferred it same as I did.
_________________
Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: M21-11  [#permalink]

### Show Tags

1
PiyushK wrote:
Bunuel wrote:
PiyushK wrote:
I think provided solution as per the language of this question is not right.

We are asked to find winning probability after the 4th throw. As per my understanding it means in sequence any throw after 4th attempt.

XXHTH
XXXXHTH
XXXXXXXHTH
XXXXXXXXXXXXHTH

I tried to find all winning cases within 4th throw.

1) HTH = P = 1/8
2) XHTH (X any turn) = P = 1/8

Probability of winning within 4 attempts = 2/8 = 1/4

Thus required probability of winning after the 4th throw = 1 - 1/4 = 3/4

The game will end after the fourth throw means that the fourth throw should be the last in the game.

Then sentence should be:
what is the probability that the game will end with 4th throw?
what is the probability that the game will end at fourth throw?

After 4th throw is ambiguous; I discussed this question with other people and they also inferred it same as I did.

Than you, Edited as you've suggested:

If a coin is tossed until the sequence $$HTH$$ appears ($$H$$ denotes the fall of heads, $$T$$ denotes the fall of tails), what is the probability that the game will end with 4th throw?
_________________
Intern  Joined: 23 Apr 2016
Posts: 21
Location: Finland
Concentration: General Management, International Business
GPA: 3.65
Re: M21-11  [#permalink]

### Show Tags

Basically it does not matter what we get in the first throw.
The last three throw should be HTH
There is only one way of getting this out of total 8 possibilities, so probability = 1/8
Manager  B
Joined: 06 Jul 2013
Posts: 59
Location: United States
GMAT 1: 720 Q49 V38 Re: M21-11  [#permalink]

### Show Tags

Bunuel wrote:
If a coin is tossed until the sequence $$HTH$$ appears ($$H$$ denotes the fall of heads, $$T$$ denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. $$\frac{1}{16}$$
B. $$\frac{3}{32}$$
C. $$\frac{1}{8}$$
D. $$\frac{3}{16}$$
E. $$\frac{1}{4}$$

Hi Bunuel or anyone else that could answer this:

given that the H in “HTH” could occur in two different ways, why isn’t the answer 1/4?

+1 for an explanation

Posted from my mobile device
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: M21-11  [#permalink]

### Show Tags

1
TippingPoint93 wrote:
Bunuel wrote:
If a coin is tossed until the sequence $$HTH$$ appears ($$H$$ denotes the fall of heads, $$T$$ denotes the fall of tails), what is the probability that the game will end with 4th throw?

A. $$\frac{1}{16}$$
B. $$\frac{3}{32}$$
C. $$\frac{1}{8}$$
D. $$\frac{3}{16}$$
E. $$\frac{1}{4}$$

Hi Bunuel or anyone else that could answer this:

given that the H in “HTH” could occur in two different ways, why isn’t the answer 1/4?

+1 for an explanation

Posted from my mobile device

What do you mean "H could occur in two different ways"? How are you getting 1/4?
_________________
Intern  B
Joined: 30 Dec 2018
Posts: 7
Re: M21-11  [#permalink]

### Show Tags

Probability of HTH in 4 chances is 1/2 * 1/2 * 1/2 * 1/2
The same can happen in 4C3 ways. = 4 * 1/16 = 1/4

Please advise where am i going wrong? Re: M21-11   [#permalink] 08 Aug 2019, 02:36
Display posts from previous: Sort by

# M21-11

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

#### MBA Resources  